EUCLIDEAN SPACES and CONVERGENCE IN R n Summary of lecture notes - - PowerPoint PPT Presentation

MAT2032 Analysis II

EUCLIDEAN SPACES and CONVERGENCE IN Rn

Summary of lecture notes mainly from Chapters 8 and 9 of our textbook: Introduction to Analysis, William R. Wade. 4th edition, Pearson, 2010. Engin MERMUT

Thanks to Sinem ODABAS ¸I and Didem COS ¸KAN who have carefully checked the typing.

Dokuz Eyl¨ ul University Faculty of Arts and Science Department of Mathematics ˙ Izmir/TURKEY

2011.03.16

Outline I

ALGEBRAIC STRUCTURE LINEAR TRANSFORMATIONS TOPOLOGY OF Rn INTERIOR, CLOSURE AND BOUNDARY LIMITS AND CONTINUITY IN TERMS OF OPEN SETS AND NEIGHBORHOODS LIMITS OF SEQUENCES IN Rn LIMITS OF FUNCTIONS CONTINUOUS FUNCTIONS COMPACTNESS AND HEINE-BOREL THEOREM CONNECTEDNESS AND PATH CONNECTEDNESS CONTINUOUS FUNCTIONS PRESERVE COMPACTNESS AND CONNECTEDNESS DINI’S THEOREM FOR MONOTONE SEQUENCES OF FUNCTIONS LEBESGUE’S CRITERION FOR RIEMANN INTEGRABILITY PARTITIONS OF UNITY

Outline II

FURTHER TOPICS: METRIC SPACES AND TOPOLOGICAL SPACES References

Euclidean spaces Rn

For each positive integer n, the n-dimensional Euclidean space is the vector space Rn = {(x1, x2, . . . , xn) | x1, x2, . . . , xn ∈ R} with its usual Euclidean inner product: For every x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) in Rn, their inner product (dot product) is defined to be x · y = x1y1 + x2y2 + · · · + xnyn =

n

• k=1

xkyk. The elements x = (x1, x2, . . . , xn) of Rn are called points or vectors or ordered n-tuples, and the numbers xj for j = 1, 2, . . . , n are called coordinates, or components of x. The zero vector is 0 = (0, 0, . . . , 0). Elements of R are called scalars. You all know the usual vector space properties of Rn that gives the algebraic structure of Rn. Throughout all these notes, whenever we write Rn, Rm, Rp, etc., n, m, p, etc. will be positive integers.

The usual orthonormal basis for the vector space Rn

The usual basis for the n-dimensional vector space Rn consists of the vectors e1, e2, . . . , en where e1 = (1, 0, 0, 0, 0, . . . , 0, 0), e2 = (0, 1, 0, 0, 0, . . . , 0, 0), e3 = (0, 0, 1, 0, 0, . . . , 0, 0), . . . = . . . en = (0, 0, 0, 0, 0, . . . , 0, 1). This usual basis is an orthonormal basis: We have for each i, j ∈ {1, 2, 3, . . . , n}, ei · ej = δij =

• if i = j,

1 if i = j.

Each vector x = (x1, x2, . . . , xn) in Rn can be expressed uniquely as a linear combination of e1, e2, . . . , en: x =

n

• k=1

xkek = x1e1 + x2e2 + · · · + xnen, and we have xk = x · ek for each k = 1, 2, . . . , n. In R2, the usual basis is denoted by {i, j}. Vectors in R2 are usually expressed in the form xi + yj where x and y are scalars (in R). In R3, the usual basis is denoted by {i, j, k}: i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1). Vectors in R3 are usually expressed in the form xi + yj + zk where x, y and z are scalars (in R).

Parallel and orthogonal vectors

Let x and y be two nonzero vectors in Rn.

◮ x and y are said to be parallel if and only if there is a scalar

t ∈ R such that x = ty.

◮ x and y are said to be orthogonal if x · y = 0.

Euclidean norm and distance in Rn

For every x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) in Rn, we define:

◮ The (Euclidean) norm (or magnitude or length) of

x = (x1, x2, . . . , xn) is defined to be the scalar x =

• x2

1 + x2 2 + · · · + x2 n =

• n
• k=1

x2

k = √x · x. ◮ The (Euclidean) distance between the two points

x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) is defined to be dist(x, y) = x − y =

• n
• k=1

(xk − yk)2. The norm makes the vector space Rn a “normed space” and this norm induces the above distance function that is a “metric” on Rn and makes Rn a “metric space”. These mean:

Properties of norm and distance

For all x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) and z = (z1, z2, . . . , zn) in Rn, and scalars α ∈ R,

• 1. x ≥ 0.
• 2. x = 0 if and only if x is the zero vector 0.
• 3. αx = |α|x.
• 4. Triangle inequality: x + y ≤ x + y (and

x − y ≥ |x − y| ≥ x − y).

• 5. dist(x, y) ≥ 0.
• 6. dist(x, y) = 0 if and only if x = y.
• 7. dist(x, y) = dist(y, x) (symmetry).
• 8. Triangle inequality: dist(x, y) ≤ dist(x, z) + dist(z, y).

You shall see metric spaces and its topology, continuous functions

• n metric spaces in your Topology course next year. You can in

advance see Chapter 10 Metric Spaces of your textbook to study the topological properties that we shall see for Rn: continuity, compactness, connectedness.

Cauchy-Schwarz Inequality

For all x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) in Rn, |x · y| ≤ x y, where equality holds if and only if one of x or y is the zero vector or they are nonzero parallel vectors (that is equality holds if and only if x and y are linearly dependent vectors in the vector space Rn). This is the Cauchy-Schwarz inequality that you have learned in your previous year (and that you shall see in your Linear Algebra course when working on inner product spaces): |x1y1+x2y2+· · ·+xnyn| ≤

• x2

1 + x2 2 + . . . + x2 n·

• y2

1 + y2 2 + . . . + y2 n,

where equality holds if and only if x is the zero vector 0 or there exists a real number u such that y = ux, that is, equality holds if and only if x1 = x2 = . . . = xn = 0 or there exists a real number u such that yk = uxk for each k = 1, 2, . . . , n.

The angle between two vectors in Rn

For all x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) in Rn, we have by Cauchy-Schwarz inequality that |x · y| ≤ x y. So if x and y are nonzero vectors, then

• x · y

x y

• ≤ 1,

that is, − 1 ≤ x · y x y ≤ 1, and so there exists a unique real number θ ∈ [0, π] such that cos θ = x · y x y and so x · y = x y cos θ. This angle θ is defined to be the angle between two nonzero vectors x and y. With this definition, x and y are parallel if and

• nly if θ = 0 or θ = π; and, x and y are orthogonal if and only if

θ = π/2.

Orthogonality and the orthogonal projection

For all vectors x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) in Rn, we have x + y2 = x2 + y2 + 2 x · y. So x and y are orthogonal, that is, x · y = 0, if and only if the Pythagorean Identity holds, that is, x + y2 = x2 + y2. If y = 0, and if we define λ = x · y y2 , then the vector z = x − λy is orthogonal to the vector y and x = z + λy, where λy is called the orthogonal projection of the vector x along the vector y: x = z + x · y y2 y = z +

• x ·

y y y y.

Note that in the last formula y y is a unit vector, that is, a vector with norm 1. If u is a unit vector in Rn, then the orthogonal projection of x along u is (x · u)u. The orthogonal projection of x along a nonzero vector y is the same with its orthogonal projection along the unit vector y y.

Cross product in the 3-dimensional space R3

The cross product of two vectors x = (x1, x2, x3) and y = (y1, y2, y3) in the 3-dimensional space R3 is the vector defined by x × y = (x2y3 − x3y2, x3y1 − x1y3, x1y2 − x2y1). This formula can be easily remembered by its formulation using determinant expansion along the first row: x × y = det   i j k x1 x2 x3 y1 y2 y3   . Using this definition, one can prove the following properties of the cross product:

Theorem

If x and y are two nonzero vectors in R3 and θ is the angle between them (that is, θ ∈ [0, π] is such that x · y = x y cos θ), then x × y = x y sin θ.

Properties of cross product in R3

Theorem

For all x = (x1, x2, x3), y = (y1, y2, y3) and z = (z1, z2, z3) in the 3-dimensional space R3, and scalars α ∈ R, we have:

• 1. x × x = 0 and y × x = −(x × y),
• 2. (αx) × y = α(x × y) = x × (αy),
• 3. x × (y + z) = (x × y) + (x × z),
• 4. (x × y) · z = x · (y × z) = det

  x1 x2 x3 y1 y2 y3 z1 z2 z3   ,

• 5. x × (y × z) = (x · z)y − (x · y)z, and
• 6. x × y2 = (x · x)(y · y) − (x · y)2 = x2y2 − (x · y)2, and

so if θ ∈ [0, π] is such that x · y = x y cos θ, then x × y2 = x2y2 − x2y2 cos2 θ = x2y2 sin2 θ.

• 7. (x × y) · x = 0 and (x × y) · y = 0. So, for two nonzero vectors

x and y that are not parallel, the vector x × y is a nonzero vector that is orthogonal to both of the vectors x and y.

Other important norms on Rn that give the same topology

For each x = (x1, x2, . . . , xn) in the n-dimensional Euclidean space Rn,

◮ The ℓ1-norm (read L-one-norm) of x = (x1, x2, . . . , xn) is the

scalar x1 = |x1| + |x2| + · · · + |xn| =

n

• k=1

|xk|.

◮ The sup-norm of x = (x1, x2, . . . , xn) is the scalar

x∞ = max({|x1|, |x2|, . . . , |xn|}). The relationship between the Euclidean norm, ℓ1-norm and sup-norm is given by the following inequalities:

Proposition

For every x = (x1, x2, . . . , xn) in Rn, we have

◮ |xk| ≤ x∞ ≤ x ≤ √nx∞ for each k = 1, 2, . . . , n, and ◮ x ≤ x1 ≤ nx∞ ≤ nx.

These inequalities are easily obtained since for each k = 1, 2, . . . , n, |xk|2 ≤ x2 = x2

1+x2 2+· · ·+x2 n ≤ n (max({|x1|, |x2|, . . . , |xn|})) = nx2 ∞,

and x2 = |x1|2+|x2|2+· · ·+|xn|2 ≤ (|x1| + |x2| + · · · + |xn|)2 = x2

1.

The important thing for all these three norms is that because of these inequalities, the “metrics” that they induce on Rn give the same collection of “open sets”, so although they are different metrics, they give Rn the same topology. These inequalities will also be useful when we need some inequalities between the coordinates of the vector x and the norm

• f the vector x.

Lines, line segments and parallelograms in Rn

Let a and b be vectors in Rn. We define:

◮ The (straight) line in Rn that passes through the point a in

the direction of the vector b = 0 is the set of points ℓa(b) = {a + tb | t ∈ R}.

◮ The line segment from the point a to the point b = a in Rn

is the set of points L(a; b) = {(1 − t)a + tb | t ∈ [0, 1]}.

◮ The entire parallelogram (with its boundary and the region it

surrounds) spanned by the vectors a and b is the set of points P(a, b) = {ua + tb | u, v ∈ [0, 1]}.

Hyperplanes in Rn

Lines in R2 and planes in R3 are generalized to hyperplanes in Rn: It will be “flat” in some sense and will have dimension n − 1. A hyperplane passing through a point a ∈ Rn with a normal b = 0 is defined to be the set Πb(a) = {x ∈ Rn | (x − a) · b = 0}, that is, it consists of all points x ∈ Rn such that x − a is

• rthogonal to b.

If b = (b1, b2, . . . , bn) and d = a · b, then the hyperplane Πb(a) consists of all x = (x1, x2, . . . , xn) ∈ Rn such that b1x1 + b2x2 + . . . bnxn = d. This gives a linear equation for the hyperplane.

Linear transformations

A function T : Rn → Rm is said to be linear if for all x, y ∈ Rn and for all scalars α ∈ R, T(x + y) = T(x) + T(y) and T(αx) = αT(x). Linear functions T : Rn → Rm are also called linear transformations or linear operators. The set of all linear transformations from Rn to Rm is denoted by L(Rn; Rm). You are studying linear transformations in your Linear Algebra

• course. You all know well operations on matrices. See also

Appendix C Matrices and Determinants (pages 629–636) of your

• textbook. We shall use elementary linear algebra as summarized

below.

Review elementary linear algebra

Note that your textbook uses the notation B = [bij]m×n to denote an m × n matrix whose (i, j)th entry, that is, the entry in the intersection of ith row and jth column is bij instead of our longer notation B = [bij]m,n

i,j=1. Anyway, what is meant is just the m × n

matrix B = [bij]m×n = [bij]m,n

i,j=1 =

     b11 b12 · · · b1n b21 b22 · · · b2n . . . . . . ... . . . bm1 bm2 · · · bmn      . For the elementary linear algebra part that we shall use be sure that you know the following:

◮ The algebra of matrices (product, sum, the m × n zero matrix

0 = 0m×n, the n × n identity matrix I = In×n, etc.).

◮ Inverse of a matrix. ◮ Properties of determinants, the expansion of a determinant

along a row or a column by minors.

◮ det(AB) = det(A) det(B) for all n × n matrices A and B.

◮ An n × n matrix is invertible if and only if det(A) = 0. ◮ The transpose of a matrix B is denoted by BT. ◮ The adjoint of an n × n matrix B is the transpose of the

matrix of cofactors of B, that is, adj(B) is the transpose of the matrix [(−1)i+j det(Bij)]n

i,j=1,

where Bij is the matrix obtained from B by deleting the ith row and jth column.

◮ If B is an invertible n × n matrix, then

B−1 = 1 det(B) adj(B).

◮ Cramer’s rule for the solution of a linear system of equations

whose coefficients matrix has nonzero determinant.

◮ For each linear transformation T : Rn → Rm, when you fix a

basis for Rn and a bases for Rm, then there is a matrix representing this linear transformation in these basis for Rn and Rm.

The matrix representing a linear transformation

Theorem

There is a one-to-one correspondence between the set L(Rn; Rm)

• f all linear transformations from Rn to Rm and the set of all

m × n matrices via the following: For each linear transformation T : Rn → Rm, there exists a unique m × n matrix B such that for each x = (x1, x2, . . . , xn) ∈ Rn, we have T(x) = y where y = (y1, y2, . . . , ym) ∈ Rm is such that B      x1 x2 . . . xn      =      y1 y2 . . . ym      . Here the m × n matrix B = [bij]m,n

i,j=1 is such that for each

j = 1, 2, . . . , n, the jth column of B is the coordinates of T(ej), where {e1, e2, . . . , en} is the usual standard basis for Rn: T(ej) = (b1j, b2j, . . . , bmj), j = 1, 2, . . . , n.

All linear transformations are of the form T(x) = Bx

By abusing notation for our usage, we shall identify points x = (x1, x2, . . . , xn) with 1 × n row matrices or n × 1 column matrices by setting [x] =

• x1

x2 · · · xn

• r its transpose

[x] =

• x1

x2 · · · xn T =      x1 x2 . . . xn      . Moreover, we shall write Bx for the product of the m × n matrix B and the n × 1 column matrix [x]. So with this agreement in the abuse of notation T(x) = Bx.

If T : Rn → Rm and U : Rm → Rp are linear transformations, then so is their composition U ◦ T : Rn → Rp. Moreover, if B is the m × n matrix that represents T and C is the p × m matrix that represents U, then the matrix product CB is the matrix that represents U ◦ T. Indeed, this is the reason why multiplication of matrices have been defined in the way you all know well. Note that with our above agreement on notation, we have: T(x) = Bx for all x ∈ Rn, U(y) = Cy for all y ∈ Rm. Thus we obtain that for all x ∈ Rn, (U ◦ T)(x) = U(T(x)) = C[T(x)] = CBx.

Note also that our identification of the points x = (x1, x2, . . . , xn) with 1 × n row matrices or n × 1 column matrices by setting [x] =

• x1

x2 · · · xn

• r its transpose

[x] =

• x1

x2 · · · xn T =      x1 x2 . . . xn      have the following properties: For all vectors x, y ∈ Rn and scalars α ∈ R,

◮ [x + y] = [x] + [y], ◮ [x · y] = [x][y]T, ◮ [αx] = α[x].

Linear transformations are needed to study differentiable functions f : Rn − → Rm

We shall use the above summarized notation for linear transformations when we study differentiable functions from Rn to

• Rm. The idea is the same as in the one variable case: The

derivative of a function f : R → R at a point a ∈ R gives the tangent line at this point and we locally approach f by the linear function g : R → R defined by g(x) = f ′(a)(x − a) + f (a), x ∈ R. The multidimensional analogue of this will replace f ′(a) (and so the linear function g) by a linear transformation T : Rn → Rm (or the matrix representing this linear transformation). The “derivative” of the function f : Rn → Rm at a point a = (a1, a2, . . . , an) ∈ Rn will be the “Jacobian matrix” of the function f : Rn → Rm, that is, the matrix of partial derivatives, evaluated at the point a = (a1, a2, . . . , an). This matrix will be called the total derivative of f at the point a.

More precisely, if for all x = (x1, x2, . . . , xn) ∈ Rn, f(x) = (f1(x), f2(x), . . . , fm(x)) ∈ Rm, then the derivative of the function f : Rn → Rm at the point a = (a1, a2, . . . , an) will be the linear transformation T : Rn → Rm that is represented by the matrix of partial derivatives evaluated at the point a = (a1, a2, . . . , an), that is, the m × n matrix ∂fi ∂xj m,n

i,j=1

=             ∂f1 ∂x1 ∂f1 ∂x2 · · · ∂f1 ∂xn ∂f2 ∂x1 ∂f2 ∂x2 · · · ∂f2 ∂xn . . . . . . ... . . . ∂fm ∂x1 ∂fm ∂x2 · · · ∂fm ∂xn             .

• f partial derivatives evaluated at the point a = (a1, a2, . . . , an).

When studying differentiability, we shall also use the operator norm

• f a linear transformation:

The operator norm of a linear transformation T : Rn → Rm

Theorem

Let T ∈ L(Rn; Rm), that is, T : Rn → Rm is a linear transformation.

• 1. There exists a real number C > 0 such that T(x) ≤ Cx

for all x ∈ Rn.

• 2. The set {C ∈ R | C > 0 and for all x ∈ Rn, T(x) ≤ Cx}
• f positive real numbers is nonempty by the first part and so

it has a greatest lower bound which is called the operator norm of T, denoted by T: T = inf({C ∈ R | C > 0 and for all x ∈ Rn, T(x) ≤ Cx}). It satisfies T(x) ≤ Tx for all x ∈ Rn. Let B be the m × n matrix that represents T so that T(x) = Bx for all x ∈ Rn. Then the operator norm B of the matrix B is also defined to be T.

• Proof. Let B be the m × n matrix that represents T so that

T(x) = Bx for all x ∈ Rn. The result is clear if B = 0. So assume that B = 0. Let b1, b2, . . . , bm be the rows of the matrix B. We then have that for all x ∈ Rn, T(x) = Bx = (b1 · x, b2 · x, . . . , bm · x). Now by Cauchy-Schwarz inequality, we obtain that T(x)2 = (b1 · x)2 + (b2 · x)2 + · · · + (bm · x)2 ≤ (b1x)2 + (b2x)2 + · · · + (bmx)2 ≤ m · [max({b12, b22, . . . , bm2})]x2 = Kx2, where K = m · [max({b12, b22, . . . , bm2})] is a finite positive real number since B = 0. So for C = √ K > 0, we have that T(x) ≤ Cx for all x ∈ Rn. Thus T is a finite nonnegative real number. By definition of T as the infimum of the set {C ∈ R | C > 0 and for all x ∈ Rn, T(x) ≤ Cx}, there exists a sequence (Ck)∞

k=1 of positive real numbers in this set

such that limk→∞ Ck = T. Thus we have that for all k ∈ Z+ and all x ∈ Rn, T(x) ≤ Ckx. Taking limit as k → ∞, we

• btain that T(x) ≤ Tx for all x ∈ Rn.

Open and closed sets in Rn

Open and closed balls and spheres are defined as follows:

◮ For a point a in Rn and a real number r > 0,

Br(a) = (open ball with center at a and radius r) = {x ∈ Rn|x − a < r}.

◮ For a point a in Rn and a real number r ≥ 0,

Br(a) = (closed ball with center at a and radius r) = {x ∈ Rn|x − a ≤ r}.

◮ For a point a in Rn+1 and a real number r > 0,

Sn

r (a)

= (n-dimensional sphere with center at a and radius r) = {x ∈ Rn+1|x − a = r}. A subset U ⊆ Rn is said to be open if for each point a in the subset U, there exists a real number r > 0 such that Br(a) ⊆ U, that is, each point a ∈ U is surrounded by an open ball lying wholly in U. A subset K ⊆ Rn is said to be closed if its complement Rn \ K is an open set in Rn.

Examples of open and closed sets

• 1. For a point a in Rn and a real number r > 0, the open ball

Br(a) is open.

• 2. For a point a in Rn and a real number r ≥ 0, the closed ball

Br(a) is closed.

• 3. For a point a in Rn+1 and a real number r > 0, the

n-dimensional sphere Sn

r (a) is closed.

• 4. For each point a ∈ Rn, the singleton {a} is closed and

Rn \ {a} is open.

• 5. The empty set ∅ and the whole space Rn are both open and

closed. Question: A set in Rn that is both open and closed is said to be a clopen set. Are there any clopen subsets of Rn except ∅ and Rn? This is indeed the question whether Rn is connected or not. See the part on connected subsets of Rn.

Union and intersection of open sets

Theorem

◮ Union of any collection of open sets (maybe an infinite

collection of open sets) in Rn is open.

◮ Intersection of finitely many open sets in Rn is open.

That is:

◮ If {Vα}α∈A is an indexed collection of open subsets of Rn,

then

• α∈A

Vα is open.

◮ If p ∈ Z+ and V1, V2, . . . , Vp are finitely many open subsets

• f Rn, then

p

• k=1

Vk = V1 ∩ V2 ∩ · · · ∩ Vp is open.

Union and intersection of closed sets

Theorem

◮ Intersection of any collection of closed sets (maybe an infinite

collection of closed sets) in Rn is closed.

◮ Union of finitely many closed sets in Rn is closed.

That is:

◮ If {Eα}α∈A is an indexed collection of closed subsets of Rn,

then

• α∈A

Eα is closed.

◮ If p ∈ Z+ and E1, E2, . . . , Ep are finitely many closed subsets

• f Rn, then

p

• k=1

Ek = E1 ∪ E2 ∪ · · · ∪ Ep is closed.

Interior, exterior and boundary points

For a set E ⊆ Rn and a point a ∈ Rn, a is said to be

◮ an interior point of E if there exists a real number r > 0

such that Br(a) ⊆ E,

◮ an exterior point of E if there exists a real number r > 0

such that Br(a) ⊆ Rn \ E,

◮ a boundary point of E if for every real number r > 0,

Br(a) ∩ E = ∅ and Br(a) ∩ (Rn \ E) = ∅. So a set E ⊆ Rn divides the whole space Rn into three disjoint pieces, two of which is open and the boundary is closed: Rn = int(E)

=E o=all interior points

• ∂E
• all boundary points
• =E=closure of E
• ext(E)

all exterior points

To prove this decomposition, we shall of course firstly define the interior and closure of a subset E of Rn.

The interior and closure of a subset E of Rn

Let E ⊆ Rn.

◮ The interior of E is the set

E o = int(E) =

• {V | V ⊆ E and V is open in Rn}.

◮ The closure of E is the set

E =

• {C | C ⊇ E and C is closed in Rn}.

Theorem

For every E ⊆ Rn, we have:

• 1. E o ⊆ E ⊆ E, E o is open and E is closed.
• 2. If V is open and V ⊆ E, then V ⊆ E o. So E o is the largest
• pen set contained in E.
• 3. If C is closed and C ⊇ E, then C ⊇ E. So E is the smallest

closed set containing E.

• 4. E is open if and only if E o = E.
• 5. E is closed if and only if E = E.

The interior, boundary and closure of a set

Now we can prove the result for a subset E ⊆ Rn that we stated at the beginning: Rn = int(E)

=E o=all interior points

• ∂E
• all boundary points
• =E=closure of E
• ext(E)

all exterior points

To prove the following theorem, firstly show that for every x ∈ Rn:

◮ x ∈ E

⇐ ⇒ Br(x) ∩ E = ∅ for all r > 0.

◮ x ∈ E o ⇐

⇒ Br(x) ∩ (Rn \ E) = ∅ for all r > 0.

Theorem

Let E ⊆ Rn. Then

• 1. E o is the set of all interior points of E and ext(E) is the set of

all interior points of Rn \ E. So ext(E) = (Rn \ E)o.

• 2. ∂E = E \ E o and E = E o ∪ ∂E.
• 3. ∂E = E ∩ (Rn \ E) and ∂E is closed.

Cluster points (= accumulation points = limit points) of a subset E ⊆ Rn

Let E ⊆ Rn. A point x ∈ Rn is said to be a cluster point (or an accumulation point or a limit point) of the set E if for every real number r > 0, the intersection Br(x) ∩ E contains a point other than the point x, that is, (Br(x) \ {x}) ∩ E = ∅ for all real numbers r > 0, equivalently, Br(x) ∩ E contains infinitely many points for all real numbers r > 0.

Theorem

The closure of a subset E ⊆ Rn is the union of E and all of its cluster points: E = E ∪ { all cluster points of E}. This holds because for a point x ∈ Rn, x ∈ E ⇐ ⇒ Br(x) ∩ E = ∅ for all r > 0, and so, x ∈ E \ E ⇐ ⇒ (Br(x) \ {x}) ∩ E = ∅ for all r > 0.

Isolated points of a subset E ⊆ Rn

Let E ⊆ Rn. Take a point a in the set E. There are two cases for the point a ∈ E:

◮ The point a ∈ E is a cluster point of E, that is,

(Br(a) \ {a}) ∩ E = ∅ for all real numbers r > 0.

◮ The point a ∈ E satisfies

(Bδ(a) \ {a}) ∩ E = ∅ for some real number δ > 0. That is, there exists an open ball with center at the point a ∈ E that contains no other point of E except of course the point a that we have taken from E: Bδ(a) ∩ E = {a} for some real number δ > 0. Such a point a ∈ E is said to be an isolated point of E; it is isolated from all other points of E since an open ball of positive radius with center at a ∈ E contains no other points

• f E.

Interior, closure and boundary of unions and intersections

Theorem

Let A and B be subsets of Rn.

• 1. (A ∪ B)o ⊇ Ao ∪ Bo,
• 2. (A ∩ B)o = Ao ∩ Bo,
• 3. A ∪ B = A ∪ B,
• 4. A ∩ B ⊆ A ∩ B,
• 5. ∂(A ∪ B) ⊆ ∂A ∪ ∂B,
• 6. ∂(A ∩ B) ⊆ (A ∩ ∂B) ∪ (B ∩ ∂A) ∪ (∂A ∩ ∂B) ⊆ ∂A ∪ ∂B.

In the above theorem, for the parts where equality need not hold, give examples where equality does not hold.

Distance of a point to a closed set

Theorem

Let E be a nonempy subset of Rn. Let a ∈ Rn. Let d be the distance of the point a to the set E defined by d = inf

x∈E x − a = inf ({x − a | x ∈ E}) . ◮ If a ∈ E, then clearly d = 0. ◮ If a ∈ E and E is closed, then d > 0.

So if E is closed, then we have that a ∈ E if and only if d = 0: Thus the points not in a closed set E are at a positive distance apart form the closed set E. This is because if E is closed and a ∈ E, then Rn \ E is open and a ∈ Rn \ E which implies the existence of a positive real number r > 0 such that Br(a) ⊆ Rn \ E, and so Br(a) ∩ E = ∅. Thus for every x ∈ E, we have that x ∈ Br(a). So for every x ∈ E, we have x − a ≥ r. This implies that d ≥ r. Since r > 0, we obtain d > 0.

Neighborhood of a point a in Rn

Let a ∈ Rn. A subset A ⊆ Rn is said to be a neighborhood of a in Rn if there exists an open set U in Rn such that a ∈ U ⊆ A, equivalently there exists a real number r > 0 such that Br(a) ⊆ A. Note that a neighborhood of a point a ∈ Rn is not necessarily an

• pen set; but it contains an open set that contains the point a. In

some textbooks, a neighborhood is assumed to be open or assumed to be an open ball. What is essential in all these definitions is that a neighborhood of a point a contains an open set that contains a. A deleted neighborhood of a point a ∈ Rn is a neighborhood of a minus the point a itself, that is, A \ {a} for a neighborhood A of a. Open sets and neighborhoods are used to state limits and continuity.

Limit of a sequence in terms of open sets

Let {xk}∞

k=1 be a sequence of points in Rn. Let a ∈ Rn. The

sequence {xk}∞

k=1 is said to converge to the point a if for every

real number ǫ > 0, there exists a positive integer N such that for all integers k, k ≥ N = ⇒ xk − a < ǫ. Notice that this last condition means that xk ∈ Bǫ(a). So we have:

Theorem

For a sequence {xk}∞

k=1 in Rn and a point a ∈ Rn, the following

are equivalent:

• 1. The sequence {xk}∞

k=1 converges to the point a.

• 2. For every open set V that contains a, there exists a positive

integer N such that for all integers k ≥ N, xk ∈ V .

• 3. For every neighborhood A of the point a, there exists a

positive integer N such that for all integers k ≥ N, xk ∈ A.

Inverse image of a function

Remember that for a function f : A → B from a set A to a set B, we define for C ⊆ A and D ⊆ B:

◮ The image of C under f is the set

f (C) = {f (x) | x ∈ C} that consists of the images of every element in the set C.

◮ The inverse image of the set D is the set

f −1(D) = {x ∈ A | f (x) ∈ D} that consists of all elements from A whose image is in the set D. Never mix this notation with the inverse of a function. The above inverse image f −1(D) is defined for every function f : A → B and D ⊆ B. But as you know well to speak about the inverse function f −1 : B → A is only possible if the function f : A → B is

• ne-to-one and onto.

Limit of a function in terms of open sets

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Let L ∈ Rm. Then f (x) is said to converge to L as x approaches a and we write lim

x→a f (x) = L if for every real number ǫ > 0, there exists a

real number δ > 0 such that for all x ∈ Rn, 0 < x − a < δ = ⇒ f (x) − L < ǫ. Notice that

◮ 0 < x − a < δ means that x ∈ Bδ(a) \ {a}, ◮ f (x) − L < ǫ means that f (x) ∈ Bǫ(L), and so ◮ ∀x ∈ Rn (0 < x − a < δ =

⇒ f (x) − L < ǫ) means that f (Bδ(a) \ {a}) ⊆ Bǫ(L) or Bδ(a) \ {a} ⊆ f −1(Bǫ(L)).

Since lim

x→a f (x) = L means that

∀ǫ > 0 ∃δ > 0 ∀x ∈ Rn (0 < x − a < δ = ⇒ f (x) − L < ǫ) , which is equivalent to ∀ǫ > 0 ∃δ > 0 Bδ(a) \ {a} ⊆ f −1(Bǫ(L)), we obtain:

Theorem

For a ∈ Rn, an open set V that contains a, L ∈ Rm and a function f : V \ {a} → Rm, the following are equivalent:

• 1. lim

x→a f (x) = L.

• 2. For every open set U in Rm that contains L, there exists an
• pen set G that contains a such that G \ {a} ⊆ f −1(U).
• 3. For every neighborhood H of L in Rm, there exists a

neighborhood A of a in Rn such that A \ {a} ⊆ f −1(H).

• 4. For every neighborhood H of L in Rm, there exists a deleted

neighborhood K of a in Rn such that K ⊆ f −1(H).

Continuity of a function at a point in terms of open sets

Let E ⊆ Rn and let f : E → Rm be a function. Let a ∈ E. The function f is said to be continuous at the point a if for every real number ǫ > 0, there exists a real number δ > 0 such that for all x ∈ Rn, x − a < δ and x ∈ E = ⇒ f (x) − f (a) < ǫ. Notice that

◮ x − a < δ means that x ∈ Bδ(a), ◮ f (x) − f (a) < ǫ means that f (x) ∈ Bǫ(f (a)), and so ◮ ∀x ∈ E (x − a < δ =

⇒ f (x) − f (a) < ǫ) means that f (E ∩ Bδ(a)) ⊆ Bǫ(f (a)) or E ∩ Bδ(a) ⊆ f −1(Bǫ(f (a))). We say that the function f : E → Rm is a continuous function if f is continuous at every point a ∈ E.

Since f : E → Rm is continuous at a point a ∈ E ⊆ Rn means that ∀ǫ > 0 ∃δ > 0 ∀x ∈ E (x − a < δ = ⇒ f (x) − f (a) < ǫ) , which is equivalent to ∀ǫ > 0 ∃δ > 0 E ∩ Bδ(a) ⊆ f −1(Bǫ(f (a))), we obtain:

Theorem

For a function f : E → Rm and a point a ∈ E ⊆ Rn, the following are equivalent:

• 1. f is continuous at the point a.
• 2. For every open set U in Rm that contains f (a), there exists an
• pen set G that contains a such that E ∩ G ⊆ f −1(U).
• 3. For every neighborhood H of f (a) in Rm, there exists a

neighborhood A of a in Rn such that E ∩ A ⊆ f −1(H).

Characterization of continuous functions in terms of open sets (and closed sets)

Theorem

For a function f : Rn → Rm, the following are equivalent:

• 1. f : Rn → Rm is a continuous function, that is, f is continuous

at every point a ∈ Rn.

• 2. f −1(V ) is an open subset of Rn for every open subset V of

Rm.

• 3. f −1(K) is a closed subset of Rn for every closed subset K of

Rm. By this theorem, continuity of a function is stated easily in terms

• f inverse images of open sets. What does happen for a function

f : E → Rm when E ⊆ Rn but E = Rn? How can we state the similar result? The answer will be to use relatively open and relatively closed subsets of E defined as follows:

Relatively open and relatively closed sets in E ⊆ Rn

Let E ⊆ Rn.

◮ A subset U ⊆ E is said to be relatively open in E if

U = V ∩ E for some open set V in Rn.

◮ A subset K ⊆ E is said to be relatively closed in E if

K = C ∩ E for some closed set C in Rn.

Theorem

For a function f : E → Rm where E ⊆ Rn, the following are equivalent:

• 1. f : E → Rm is a continuous function, that is, f is continuous

at every point a ∈ E.

• 2. f −1(V ) is a relatively open subset of E for every open subset

V of Rm.

• 3. f −1(K) is a relatively closed subset of E for every closed

subset K of Rm.

Theorem

Let E ⊆ Rn and let U ⊆ E .

• 1. U is relatively open in E if and only if for each a ∈ U, there

exists a real number r > 0 such that Br(a) ∩ E ⊆ U.

• 2. If E is an open subset of Rn, then U is relatively open in E if

and only if U is open in Rn.

• 3. If E is a closed subset of Rn, then U is relatively closed in E if

and only if U is closed in Rn.

Characterization of continuity of a function f : E → F in terms of relatively open and relatively closed sets

Theorem

For a function f : E → F where E ⊆ Rn and F ⊆ Rm, the following are equivalent:

• 1. f : E → F is a continuous function, that is, f is continuous at

every point a ∈ E.

• 2. f −1(V ) is a relatively open subset of E for every relatively
• pen subset V of F.
• 3. f −1(K) is a relatively closed subset of E for every relatively

closed subset K of F.

Restriction of continuous functions are continuous

Theorem

Let f : E → Rm be a function where E ⊆ Rn. Let H ⊆ E and let Y ⊆ Rm be such that f (H) ⊆ Y . Let g : H → Y be the restriction of the function f obtained by restricting the domain of f to H and the range of f to Y : g : H → Y defined by g(x) = f (x) for all x ∈ H. We can define g in this way since f (H) ⊆ Y . If f : E → Rm is a continuous function, then its restriction g : H → Y is also a continuous function.

Closure of a set E ⊆ Rn in terms of convergent sequences in E

Theorem

Let E ⊆ Rn.

• 1. The closure of E consists of limits of all convergent sequences

in E: E =

• lim

k→∞ xk | {xk}∞ k=1 is a convergent sequence in E

• .
• 2. E is closed if and only E contains all the limits of convergent

sequences in E, that is, for every convergent sequence {xk}∞

k=1 in E, the limit lim k→∞ xk is also in E.

Sequences in Rn

Let {xk}∞

k=1 be a sequence of points in Rn. ◮ Let a ∈ Rn. The sequence {xk}∞ k=1 is said to converge to

the point a if for every real number ǫ > 0, there exists a positive integer N such that for all integers k, k ≥ N = ⇒ xk − a < ǫ. In this case, we write xk → a as k → ∞, or, lim

k→∞ xk = a and

say that a is the limit of the sequence {xk}∞

k=1. ◮ {xk}∞ k=1 is said to be a convergent sequence if there exists

a ∈ Rn such that the sequence {xk}∞

k=1 converges to a. ◮ {xk}∞ k=1 is said to be a Cauchy sequence if for every real

number ǫ > 0, there exists N ∈ Z+ such that for all integers k and r, k ≥ N and r ≥ N = ⇒ xk − xr < ǫ.

◮ {xk}∞ k=1 is said to be a bounded sequence if there exists a

real number M > 0 such that xk ≤ M for all k ∈ Z+.

Limits are taken in each coordinate separately

To denote the coordinates of the elements of the sequence {xk}∞

k=1 in Rn, let us write for each k ∈ Z+,

xk = (xk(1), xk(2), . . . , xk(n)) ∈ Rn. With this notation:

Theorem

Let {xk}∞

k=1 be a sequence in Rn and let a ∈ Rn whose coordinates

are given by a = (a(1), a(2), . . . , a(n)) ∈ Rn and for all k ∈ Z+, xk = (xk(1), xk(2), . . . , xk(n)) ∈ Rn. Then the sequence {xk}∞

k=1 converges to the point a if and only if

for each j = 1, 2, . . . , n, the component sequence {xk(j)} of real numbers converges to the real number a(j), that is, the sequence

• f jth coordinates of the sequence {xk}∞

k=1 converges to the jth

coordinate of a for each j = 1, 2, . . . , n: lim

k→∞ xk = a

⇐ ⇒ lim

k→∞ xk(j) = a(j) for each j = 1, 2, . . . , n.

The countable set Qn is dense in Rn

The subset Q of all rational numbers is dense in R, that is, between any two real numbers, there exists a rational number. Using this, for any real number a, we can find a sequence of rational numbers that converges to a. Doing this componentwise for a point a ∈ Rn, we obtain:

Theorem

For each a ∈ Rn, there is a sequence {xk}∞

k=1 in Qn such that

lim

k→∞ xk = a.

This means that each point of Rn is in the closure of the set Qn. Such sets are said to be dense in Rn. A subset A of Rn is said to be dense in Rn if A = Rn . Since the set Qn of all points whose coordinates are rational numbers is also a countable set, we have found a countable dense subset of Rn which is important for the structure of Rn as we shall see in the proof of Lindel¨

• f theorem.

Properties of limits of sequences in Rn

The following results are proved like in the case of sequences of real numbers mainly by just replacing the absolute value |x − a| for real numbers x and a with x − a for points x, a ∈ Rn.

◮ A sequence in Rn can have at most one limit, that is, if a

sequence {xk}∞

k=1 converges to a point a and to a point b,

then we must have a = b. So we can speak of the limit of a convergent sequence in Rn.

◮ If a sequence {xk}∞ k=1 in Rn converges to a point a ∈ Rn and

if {xkj}∞

j=1 is a subsequence of {xk}∞ k=1, then we have

lim

j→∞ xkj = a. ◮ Every convergent sequence in Rn is bounded.

If {xk}∞

k=1 and {yk}∞ k=1 are convergent sequences in Rn and

α ∈ R, then

◮

lim

k→∞ (xk + yk) = lim k→∞ xk + lim k→∞ yk, ◮

lim

k→∞ (αxk) = α lim k→∞ xk, ◮

lim

k→∞ (xk · yk) =

• lim

k→∞ xk

• ·
• lim

k→∞ yk

• ,

◮

lim

k→∞ xk =

• lim

k→∞ xk

• , and

◮ moreover, if n = 3, we have

lim

k→∞ (xk × yk) =

• lim

k→∞ xk

• ×
• lim

k→∞ yk

• .

Bolzano-Weirstrass Theorem and Completeness of Rn

Theorem

Bolzano-Weirstrass Theorem for Rn. Every bounded sequence in Rn has a convergent subsequence. To prove Bolzano-Weirstrass Theorem for Rn, apply Bolzano-Weirstrass Theorem for sequences of real numbers to each coordinate separately.

Theorem

A sequence in Rn is a Cauchy sequence if and only if it is a convergent sequence. It is easily seen that every convergent sequence in Rn is a Cauchy

• sequence. To prove the harder part that every Cauchy sequence in

Rn is convergent, follow the proof as in the case of real numbers. Firstly, show that every Cauchy sequence {xk}∞

k=1 in Rn is

bounded and so by Bolzano-Weirstrass Theorem for Rn has a convergent subsequence with limit a ∈ Rn say. Now show that the whole Cauchy sequence {xk}∞

k=1 must also converge to a.

Vector valued functions of a vector variable

A vector valued function of a vector variable is a function f : A → Rm for some A ⊆ Rn, where m, n ∈ Z+. Thus it is a function from n variables to m variables, that is, for each x = (x1, x2, . . . , xn) ∈ A ⊆ Rn, there exists a unique y = (y1, y2, . . . , ym) ∈ Rm such that f (x) = y, that is, f (x1, x2, . . . , xn) = (y1, y2, . . . , ym). So there exist m functions f1, f2, . . . , fm such that fk : A → R for each k = 1, 2, . . . , m and for each x = (x1, x2, . . . , xn) ∈ A ⊆ Rn, there exists a unique y = (y1, y2, . . . , ym) ∈ Rm such that f (x) = y = (y1, y2, . . . , ym) = (f1(x), f2(x), . . . , fm(x)), that is, fk(x) = fk(x1, x2, . . . , xn) = yk for each k = 1, 2, . . . , m. Thus we have for each x = (x1, x2, . . . , xn) ∈ A ⊆ Rn, f (x) = (f1(x), f2(x), . . . , fm(x)), that is, f (x1, x2, . . . , xn) = (f1(x1, x2, . . . , xn), . . . , fm(x1, x2, . . . , xn)).

Vector functions f (x) = (f1(x), f2(x), . . . , fm(x)) ∈ Rm, x ∈ A ⊆ Rn

So for any function f : A → Rm where A ⊆ Rn, we have the so called component functions or coordinate functions of the function f ; they are the functions f1, f2, . . . , fm from A to R such that f (x) = (f1(x), f2(x), . . . , fm(x)) for all x ∈ A ⊆ Rn. In this case, we simply write f = (f1, f2, . . . , fm). So giving a function f : A → Rm where A ⊆ Rn means giving m real-valued functions from A. Thus if you have m functions f1, f2, . . . , fm from a set A ⊆ Rn to the set R of real numbers, then you can define a function f : A → Rm by defining f (x) = (f1(x), f2(x), . . . , fm(x)) for all x ∈ A ⊆ Rn.

Maximal domain of a function f = (f1, f2, . . . , fm)

When you are given m real-valued functions f1, f2, . . . , fm of n real variables, that is, for example, you are given some m formulas f1(x1, x2, . . . , xn), f2(x1, x2, . . . , xn), . . . , fm(x1, x2, . . . , xn) in terms of n variables x1, x2, . . . , xn, then you can define a function f by setting f (x1, x2, . . . , xn) = (f1(x1, x2, . . . , xn), . . . , fm(x1, x2, . . . , xn)) at every (x1, x2, . . . , xn) ∈ Rn where the m formulas f1(x1, x2, . . . , xn), f2(x1, x2, . . . , xn), . . . , fm(x1, x2, . . . , xn) are all

• defined. The set A consisting of all x = (x1, x2, . . . , xn) ∈ Rn such

that f1(x1, x2, . . . , xn), f2(x1, x2, . . . , xn), . . . , fm(x1, x2, . . . , xn) are all defined is said to be the maximal domain of the function f = (f1, f2, . . . , fm). Simply, domain(f ) = domain(f1) ∩ domain(f2) ∩ · · · ∩ domain(fm).

Operations on vector functions: αf , f + g, f · g, . . .

Let f : A → Rm and g : A → Rm be functions where A ⊆ Rn. Let α ∈ R. We define the following functions:

◮ αf : A → Rm is the function defined by (αf )(x) = αf (x) for

all x ∈ Rn. It is called the scalar product of α ∈ R with the function f .

◮ f + g : A → Rm is the function defined by

(f + g)(x) = f (x) + g(x) for all x ∈ Rn. It is called the sum

• f the functions f and g.

◮ f · g : A → R is the real-valued function defined by

(f · g)(x) = f (x) · g(x) for all x ∈ Rn. It is called the Euclidean dot product of the functions f and g.

◮ When m = 3, f × g : A → R3 is the function defined by

(f × g)(x) = f (x) × g(x) for all x ∈ Rn. It is called the cross product of the functions f and g.

◮ When m = 1 and g(x) = 0 for all x ∈ A, then the function

f /g = f g : A → R is defined by f g

• (x) = f (x)

g(x) for all x ∈ A.

Composition of functions

Let f : A → Rm and g : B → Rp be functions where A ⊆ Rn and B ⊆ Rm. Then the composition g ◦ f is defined at all points x ∈ A such that f (x) ∈ B by (g ◦ f )(x) = g(f (x)). That is, the domain of the function g ◦ f is A ∩ f −1(B): g ◦ f : A ∩ f −1(B) → Rp is defined by (g ◦ f )(x) = g(f (x)) for all x ∈ A ∩ f −1(B).

Limit of a function

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Let L ∈ Rm. Then f (x) is said to converge to L as x approaches a if for every real number ǫ > 0, there exists a real number δ > 0 such that for all x ∈ Rn, 0 < x − a < δ = ⇒ f (x) − L < ǫ. In this case, we write lim

x→a f (x) = L and say that L is the limit of

f (x) as x approaches a. Like in the case of real-valued functions of a real variable, the following properties of limits are obtained.

Sequential characterization of limits

Theorem

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Let L ∈ Rm. Then lim

x→a f (x) = L if and only if lim k→∞ f (xk) = L for every sequence

{xk}∞

k=1 in V \ {a} that converges to a.

Limit of sum, product, etc. of functions

Theorem

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm and g : V \ {a} → Rm be functions. Let α ∈ R. If f (x) and g(x) have limits as x approaches a, then (f + g)(x), (αf )(x), (f · g)(x) and f (x) also have limits as x approaches a and they satisfy

◮ lim x→a(f + g)(x) = lim x→a f (x) + lim x→a g(x), ◮ lim x→a(αf )(x) = α lim x→a f (x), ◮ lim x→a(f · g)(x) =

• lim

x→a f (x)

• ·
• lim

x→a g(x)

• ,

◮

• lim

x→a f (x)

• = lim

x→a f (x).

◮ Moreover, when m = 3, (f × g)(x) has a limit as x

approaches a and satisfies lim

x→a(f × g)(x) =

• lim

x→a f (x)

• ×
• lim

x→a g(x)

• .

◮ If m = 1 and lim x→a g(x) = 0, then

f g

• (x) = f (x)

g(x) has a limit as x approaches a and satisfies lim

x→a

f (x) g(x)

• =

lim

x→a f (x)

lim

x→a g(x).

Sequeeze Theorem for functions

Theorem

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → R, g : V \ {a} → R and h : V \ {a} → R be real-valued functions such that f (x) ≤ g(x) ≤ h(x) for all x ∈ V \ {a}. If there exists a real number L such that lim

x→a f (x) = lim x→a h(x) = L,

then g(x) also has a limit as x approaches a and satisfies lim

x→a g(x) = L.

Limit of composition of functions

Theorem

Let a ∈ Rn and let V be an open set that contains a. Let L ∈ Rm and let U be an open set that contains L. Let f : V \ {a} → Rm and g : U → Rp be functions. If lim

x→a f (x) = L and g is continuous at L, then

lim

x→a(g ◦ f )(x) = g(L),

that is, we have in this case, lim

x→a g(f (x)) = g

• lim

x→a f (x)

• .

Limits are found in each coordinate separately

Theorem

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Let L = (L1, L2, . . . , Lm) ∈ Rm. Let f1, f2, . . . , fm be the component functions of f , that is, f = (f1, f2, . . . , fm) where for each j = 1, 2, . . . , m, fj is the real-valued function defined on V \ {a} such that f (x) = (f1(x), f2(x), . . . , fm(x)) for all x ∈ V \ {a}. Then L = lim

x→a f (x) if and only if

Lj = lim

x→a fj(x)

for each j = 1, 2, . . . , m. By this theorem, finding the limit of vector-valued functions of a vector variable reduces to finding the limits of real-valued functions

• f a vector variable.

To prove a limit

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Let L ∈ Rm. To prove that L = lim

x→a f (x) what we usually do is to dominate

f (x) − L by a nonnegative real-valued function g : V \ {a} → R such that lim

x→a g(x) = 0:

f (x) − L ≤ g(x) and lim

x→a g(x) = 0.

lim

x→a f (x) = L implies f (x) → L as x → a through all paths

Theorem

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Suppose that lim

x→a f (x) exists, say for some L ∈ Rm,

L = lim

x→a f (x).

Then for every function g : I → V \ {a} such that I is an open interval in R that contains 0 and lim

t→0 g(t) = a,

we must also have lim

t→0 f (g(t)) = L.

This theorem means that if lim

x→a f (x) = L exists, then f (x) → L as

x → a over any path, that is, the limit value must be the same through all paths that x approaches a. Using this theorem, we may be able to prove that lim

x→a f (x) does

not exist by finding two different paths that x approaches a such that the limits over these paths are not equal.

To prove that a limit does not exist

Let a ∈ Rn and let V be an open set that contains a. Let f : V \ {a} → Rm be a function. Suppose we wish to prove that lim

x→a f (x)

does not exist. We look for two paths g1 : I → V \ {a} and g2 : I → V \ {a} such that I is an open interval in R that contains 0 and lim

t→0 g1(t) = a

and lim

t→0 g2(t) = a,

but lim

t→0 f (g1(t)) = lim t→0 f (g2(t)).

Then according to the above theorem, it is not possible that lim

x→a f (x) exists.

Do not be mislead by this method. It gives the result that lim

x→a f (x)

does not exist only if you can find two such paths g1 and g2 as described above. For example, it may be that over paths which are lines, the limit values are always the same. This does not mean that the limit lim

x→a f (x) exists. This is only a method to prove that

a limit does not exist if you are able to find two different paths approaching a such that f (x) has different limits along these

• paths. To prove that a limit exists, you must directly use the

definition and the results for known limits (and the known continuous functions). Solve some examples by using this method to prove that a limit does not exist.

Iterated limits

Theorem

Let I and J be open intervals in R, let a ∈ I and let b ∈ J. Let f : (I × J) \ {(a, b)} → R be a function. Assume that the following three conditions hold:

• 1. lim

y→b f (x, y) exists for for each x ∈ I \ {a},

• 2. lim

x→a f (x, y) exists for for each y ∈ J \ {b},

3. lim

(x,y)→(a,b) f (x, y) exists, say for some L ∈ R,

lim

(x,y)→(a,b) f (x, y) = L.

Then both of the iterated limits lim

x→a lim y→b f (x, y)

and lim

y→b lim x→a f (x, y)

exist and they satisfy lim

x→a lim y→b f (x, y) = lim y→b lim x→a f (x, y) = L.

This theorem on iterated limits cannot be strengthened in general. Firstly, the iterated limits of a given function f (x, y) may not exist. When the iterated limits exist, they may not be equal. Even when the iterated limits exist and are equal, it may be that the limit lim

(x,y)→(a,b) f (x, y) does not exist. Give examples for all these cases.

Continuity and uniform continuity of a function

Let E ⊆ Rn and let f : E → Rm be a function.

◮ Let a ∈ E. The function f is said to be continuous at the

point a if for every real number ǫ > 0, there exist a real number δ > 0 such that for all x ∈ Rn, x − a < δ and x ∈ E = ⇒ f (x) − f (a) < ǫ.

◮ Let B ⊆ E. The function f : E → Rm is said to be

continuous on B if f is continuous at every point a ∈ B.

◮ The function f : E → Rm is said to be continuous if it is

continuous on its domain E.

◮ Let B ⊆ E. The function f : E → Rm is said to be uniformly

continuous on B if for every real number ǫ > 0, there exists a real number δ > 0 such that for all x ∈ Rn and for all a ∈ Rn, x−a < δ and x ∈ B and a ∈ B = ⇒ f (x)−f (a) < ǫ.

◮ The function f : E → Rm is said to be uniformly continuous

if it is uniformly continuous on its domain E.

Sequential characterization of continuity

Theorem

Let E ⊆ Rn and let f : E → Rm be a function. Let a ∈ E. The function f is continuous at the point a if and only if lim

k→∞ f (xk) = f (a)

for every sequence {xk}∞

k=1 in E that converges to a.

Sequential characterization of uniform continuity

Theorem

Let E ⊆ Rn and let f : E → Rm be a function. The function f is uniformly continuous (on E) if and only if for all sequences {xk}∞

k=1 and {yk}∞ k=1 in E,

lim

k→∞ xk − yk = 0

= ⇒ lim

k→∞ f (xk) − f (yk) = 0.

Continuity of sum, product, etc. of functions

Theorem

Let E ⊆ Rn and a ∈ E. Let f : E → Rm and g : E → Rm be functions. Let α ∈ R. Assume that f and g are continuous at a. Then:

◮ αf , f · g and f are also continuous at a. ◮ If m = 3, f × g is also continuous at a. ◮ If m = 1 and g(x) = 0 for all x ∈ E, then f /g = f

g : E → R is also continuous at a.

Theorem

Let E ⊆ Rn. Let f : E → Rm and g : E → Rm be functions. Let α ∈ R. If f and g are continuous functions, then the following functions are also continuous functions:

◮ αf , f · g and f , ◮ f × g when m = 3, ◮ f /g = f

g : E → R when m = 1 and g(x) = 0 for all x ∈ E.

Composition of continuous functions is continuous

Theorem

Let E ⊆ Rn and ∅ = G ⊆ Rm. Let f : E → Rm and g : G → Rp be functions such that f (E) ⊆ G so that the composition function g ◦ f : E → Rp is defined by (g ◦ f )(x) = g(f (x)) for all x ∈ E.

◮ If f is continuous at a point a ∈ E and g is continuous at

f (a), then g ◦ f is also continuous at a.

◮ If f and g are continuous functions, then their composition

g ◦ f is also continuous. Prove this theorem using the characterization of continuity in terms of open sets.

Continuity in terms of component functions

Theorem

Let E ⊆ Rn and f : E → Rm be a function. Let f1, f2, . . . , fm be the component functions of f , that is, f = (f1, f2, . . . , fm) where for each j = 1, 2, . . . , m, fj is the real-valued function defined on E such that f (x) = (f1(x), f2(x), . . . , fm(x)) for all x ∈ E. Then

◮ f is continuous at a point a ∈ E if and only if fj is continuous

at a for each j = 1, 2, . . . , m.

◮ f is a continuous function if and only if all of the component

functions fj are continuous for j = 1, 2, . . . , m. By this theorem, to determine the continuity of a vector-valued function of a vector variable, you just need to look at the continuity of its component functions which are real-valued functions of a vector variable.

Examples of continuous functions

◮ The identity function f : Rn → Rn defined by f (x) = x,

x ∈ Rn is continuous and so its component functions are also continuous, that is, for each j = 1, 2, . . . , n, the function fj : Rn → R defined by fj(x) = xj for every x = (x1, x2, . . . , xn) ∈ Rn is continuous. The function fj : Rn → R is called the jth projection function.

◮ Every linear transformation T : Rn → Rm is continuous.

Prove this using the norm of a linear operator or prove it directly by considering its component functions.

◮ Using the norm of a linear operator, you can easily prove that

every linear transformation T : Rn → Rm is indeed uniformly continuous.

◮ Polynomial functions of n variables are continuous. A

polynomial function of n variables x1, x2, . . . , xn is a function P : Rn → R defined for all x = (x1, x2, . . . , xn) ∈ Rn by P(x) = P(x1, x2, . . . , xn) =

p1

• k1=0

p2

• k2=0

· · ·

pn

• kn=0

ck1k2···knxk1

1 xk2 2 · · · xkn n .

Here p1, p2, . . . , pn are some fixed nonnegative integers and ck1k2···kn is a constant real number for each n-tuple (k1, k2, . . . , kn) of nonnegative integers such that kj ≤ pj for every j = 1, 2, . . . , n.

◮ Rational functions of n variables are continuous. A rational

function f : Rn → R is a function defined by the quotient of two polynomial functions, that is, for some polynomial functions P : Rn → R and Q : Rn → R, f (x) = P(x) Q(x) for all x ∈ Rn such that Q(x) = 0. In this case, the rational function is continuous on its domain, that is, on the set {x ∈ Rn | Q(x) = 0}.

◮ The functions produced by composing continuous functions

are continuous and so many continuous functions are produced using the basic functions above and the continuous functions that you have seen in your calculus courses. For example, the function f : R2 → R3 defined by f (x, y) = (x2 sin y, ex ln(1+y2), x3−5 cos y), (x, y) ∈ R2, is continuous.

Sign-preserving property of real-valued continuous functions

Theorem

Let E ⊆ Rn and a ∈ E. Let f : E → R be a real-valued function defined on the subset E ⊆ Rn.

◮ If f is continuous at a and f (a) > 0, then there exist positive

real numbers L and δ such that for all x ∈ Rn, x − a < δ and x ∈ E = ⇒ f (x) > L. Indeed, one can take for example L = f (a) 2 .

◮ If f is continuous at a and f (a) < 0, then there exist positive

real numbers L and δ such that for all x ∈ Rn, x − a < δ and x ∈ E = ⇒ f (x) < −L. Indeed, one can take for example L = −f (a) 2 .

Theorem

Let E ⊆ Rn and a ∈ E. Let f : E → R be a real-valued function defined on the subset E ⊆ Rn. If f is continuous at a and f (a) = 0, then there exist positive real numbers L and δ such that for all x ∈ Rn, x − a < δ and x ∈ E = ⇒ |f (x)| > L. Indeed, one can take for example L = |f (a)| 2 .

Continuity of the quotient function at a point where the denominator is nonzero

Theorem

Let E ⊆ Rn and a ∈ E. Let f : E → R and g : E → R be real-valued functions. Assume that f and g are continuous at a, and moreover, g(a) = 0. Then there exists a real number δ > 0 such that for all x ∈ E ∩ Bδ(a), g(x) = 0 and so the quotient function f /g = f g : E ∩ Bδ(a) → R is well-defined on E ∩ Bδ(a) and it is continuous at a.

Open and closed sets defined by continuous functions

For any real number c, the intervals (c, ∞) and (−∞, c) are open in R and the intervals [c, ∞) and (−∞, c] are closed in R. So using the characterization of continuity in terms of open and closed sets, we obtain:

Theorem

Let f : Rn − → R be a continuous function and let c be a real

• number. Then:
• 1. The sets {x ∈ Rn | f (x) < c}, {x ∈ Rn | f (x) > c} and

{x ∈ Rn | f (x) = c} are open in Rn.

• 2. The sets {x ∈ Rn | f (x) ≤ c}, {x ∈ Rn | f (x) ≥ c} and

{x ∈ Rn | f (x) = c} are closed in Rn. For example, by this theorem:

• 1. {(x, y, z) ∈ R3 | 2x + 3y + 4z > 0} is open in R3.
• 2. {(x, y, z) ∈ R3 | x2 + y2 + z2 = 1} is closed in R3.

Collection of sets

Let V be a collection of sets, that is, every element of the set V is also a set. The union of all the sets in the collection V is the set

• V = {x | x ∈ V for some V ∈ V}.

It may also be written as

• V ∈V

V =

• V = {x | x ∈ V for some V ∈ V}.

It may be that this collection V of sets is given in the form V = {Vα | α ∈ A} for some set A. Be careful, we do not assume that for every α = β in A, the set Vα = Vβ. In this case, the union of the sets in the collection V = {Vα | α ∈ A} is written also in the following form:

• V =
• α∈A

Vα = {x | x ∈ Vα for some α ∈ A}.

When the set A = {1, 2, . . . , n} is a finite set, we write

• V =
• α∈A

Vα =

n

• k=1

Vk. When the set A = Z+ is the set of all positive integers, we write

• V =
• α∈A

Vα =

∞

• k=1

Vk. Indeed, any collection V of sets can be written in the form V = {Vα | α ∈ A} by taking A = V and Vα = α for every α ∈ A = V: V = {Vα | α ∈ A} = {V | V ∈ V}.

Family of sets and indexed collection of sets

Let X be a set and suppose we have a function f : A → P(X) from the set A to the power set P(X) of X, that is, to the set of all subsets of X. Then for each α ∈ A, by taking Vα = f (α), we

• btain that the image of the function f is

{Vα | α ∈ A}. If we do not deal with the function f , we just write {Vα}α∈A to mean the function f that sends each α ∈ A to Vα and call {Vα}α∈A an indexed family of subsets of X. The set A is called the index set of this family. But be careful, we do not assume that the function f is one-to-one, that is, we do not assume that Vα = Vβ for every α = β in A. This is like considering sequences in a set X: A sequence {xn}∞

n=1

in a set X is nothing but just a function f : Z+ → X given by f (n) = xn for every n ∈ Z+. But you do not use function notation f : Z+ → X to denote a sequence; you just write “a sequence {xn}∞

n=1 in a set X”. The better notation not to mix sequences

with sets is to write (xn)∞

n=1.

Similarly, a better notation for an indexed family of subsets of X will be to write (Vα)α∈A. The collection {Vα | α ∈ A} is the image

• f the function f and it is a subset of the power set P(X) of X.

But by (Vα)α∈A, we mean the function f : A → P(X) such that f (α) = Vα for each α ∈ A. These details may not be important for your course now, but be careful that there is indeed a difference. You can forget these details between an indexed family {Vα}α∈A of subsets of a set X and the set {Vα | α ∈ A} that is a collection of

• sets. Because, what we use now is just writing a collection V of

subsets of X in the form V = {Vα | α ∈ A} for some set A. And we write this even more shorter by saying that {Vα}α∈A is an (indexed) collection of subsets of X.

Covering of a set, finite subcovering and countable subcovering

Let E ⊆ Rn. Let {Vα}α∈A be a collection of subsets of Rn.

◮ {Vα}α∈A is said to be a cover of E (or a covering of E) if

• α∈A

Vα ⊇ E. In this case, we also say that {Vα}α∈A covers E.

◮ {Vα}α∈A is said to be an open cover of E (or an open

covering of E) if it is a cover of E such that Vα is open for every α ∈ A.

◮ A cover {Vα}α∈A of E is said to have a finite subcover (a

finite subcovering) if there exists a finite subset B of the index set A such that {Vα}α∈B also covers E.

◮ A cover {Vα}α∈A of E is said to have a countable subcover

(countable subcovering) if there exists a countable subset B

• f the index set A such that {Vα}α∈B also covers E.

Definition of compact subsets of Rn

Let E ⊆ Rn. We say that E is a compact set if EVERY open covering of E has a finite subcover. See the lecture notes (or your textbook) for the proof of the following theorem:

Theorem

Let E ⊆ Rn and f : E → Rm be a function. If E is a compact set and f : E → Rm is a continuous function, then f : E → Rm is uniformly continuous. SEE ALSO MY HANDWRITTEN NOTES FOR THE TOPOLOGY OF R, COMPACTNESS IN R AND UNIFORM CONTINUITY OF REAL-VALUED FUNCTIONS OF A REAL VARIABLE. UNDERSTANDING COMPACTNESS AND UNIFORM CONTINUITY IN R WILL BE VERY

• HELPFUL. STUDY THESE HANDWRITTEN NOTES

ALSO FOR A REVIEW OF CONTINUITY AND LIMITS OF SEQUENCES IN R.

Compact subsets of Rn are always closed and bounded

A subset A of Rn is said to be bounded if there exists a real number M > 0 such that for all x ∈ A, x < M. Unbounded means of course not bounded. Using directly the definition of compact sets, the following properties of compact subsets of Rn are proved.

Theorem

◮ The empty set and all finite subsets of Rn are compact. ◮ Compact subsets of Rn are necessarily closed. ◮ Compact subsets of Rn are necessarily bounded. ◮ A closed subset of a compact subset of Rn is compact.

What about the converse? Is every closed and bounded subset of Rn compact?

Proof: The first result is easily proved. The second and third part say that every compact subset of Rn is closed and bounded. This is the easier part of Heine-Borel theorem; they are proved in the proof of Heine-Borel theorem

• below. The converse also holds and this is Heine-Borel theorem.

Let’s prove the last part. Let H be a compact subset of Rn and let K ⊆ H be such that K is a closed set in Rn. We shall show that K is compact, too. Let {Vα}α∈A be an open covering of K. Since K is closed, Rn \ K is open. Adding the open set Rn \ K to the {Vα}α∈A, we obtain an open covering {Vα | α ∈ A} ∪ {Rn \ K} of

• H. Since H is compact, this open covering has a finite subcover.

So we can assume that for some finite set B ⊆ A, {Vα | α ∈ B} ∪ {Rn \ K} covers H. Then necessarily {Vα}α∈B must cover K ⊆ H. Thus every open covering of K has a finite

• subcover. By definition of compactness, this means that K is

compact.

Heine-Borel Theorem in Rn: compact = closed + bounded

Theorem

Heine-Borel Theorem. A subset of Rn is compact if and only if it is closed and bounded. As we have said before every compact set is always closed and

• proved. What remains to be proved in Heine-Borel theorem is to

prove that every closed and bounded subset of Rn is compact. There are a few ways to prove this. For the proofs of Heine-Borel theorem in R (that is for n = 1), see my handwritten notes for the topology of R. The proof in your textbook uses Lindel¨

• f Theorem

and Borel Covering Lemma. We shall also give the outline for another proof. A third proof uses the bisection method, this is the method that we have used to prove Bolzano-Weirstrass theorem for sequences of real numbers. The relation with Bolzano-Weirstrass theorem is not surprising since we have the following theorem whose proof is left to you as an exercise:

Theorem

For a subset E of Rn, the following are equivalent:

• 1. E is compact.
• 2. E is closed and bounded.
• 3. Every infinite subset of E has a cluster point in E.
• 4. Every sequence in E has a convergent subsequence whose

limit is in E. If a set E ⊆ Rn satisfies the last condition of the above theorem, then E is said to be sequentially compact. Thus the theorem gives us that for a subset E ⊆ Rn, being compact and being sequentially compact are equivalent. You have seen Bolzano-Weirstrass Theorem for sequences in Rn: Every bounded sequence in Rn has a convergent subsequence. Similarly, prove:

Theorem

Bolzano-Weirstrass Theorem for sets: Every infinite bounded subset E of Rn has a cluster point in Rn.

Lindel¨

• f: Every open covering has a countable subcovering

If a subset E ⊆ Rn is compact, then every open covering of E has a finite subcover. That is the definition of compactness. But whether E is compact or not, it is always true that every open covering of E has a countable subcovering:

Theorem

Lindel¨

• f Covering Theorem. Every open covering of a subset of

Rn has a countable subcovering. That is, if E ⊆ Rn and {Vα}α∈A is a collection of open sets in Rn such that

• α∈A

Vα ⊇ E, then there exists a countable subset B of A such that

• α∈B

Vα ⊇ E.

The proof of Lindel¨

• f Covering Theorem uses the countable dense

subset Qn of Rn. The collection U consisting of all open balls whose centers are in Qn and whose radius is a rational number is a countable collection of open balls and has the following property: For a subset A of Rn, A is open if and only if for every a ∈ A, there exists an element U ∈ U such that a ∈ U ⊆ A, that is, U is an open ball whose center have all rational coordinates and whose radius is a rational number, and that U lies wholly in A and U contains the point a. This is easily proved because for a point a = (a1, a2, . . . , an) ∈ Rn and real number r > 0, we choose a point b = (b1, b2, . . . , bn) in the open ball Br(a) sufficiently near to the point a = (a1, a2, . . . , an) such that all of the coordinates of the point b = (b1, b2, . . . , bn) are rational numbers and there exists a rational number q such that the open ball Bq(b) contains the point a and this open ball Bq(b) is contained in the open ball Br(a): a ∈ Bq(b) ⊆ Br(a).

This is obtained using the fact that the set Q is dense in R. The rational numbers b1, b2, . . . , bn are chosen such that |bk − ak| ≤ r 4n for each k = 1, 2, . . . , n, and the rational number q is chosen such that r 4 < q < r

• 2. Prove

that with this choice of rational numbers b1, b2, . . . , bn and q, we have a ∈ Bq(b) ⊆ Br(a), that is, prove that for every x ∈ Rn, x − b < q = ⇒ x − a < r. Since the collection U of all open balls whose centers are in Qn and whose radius is a rational number is a countable collection of

• pen balls, you can assume that its elements are indexed by the set

Z+ of positive integers, that is, say U = {Uk | k ∈ Z+}. Using the above property of this countable collection U of open balls, prove Lindel¨

• f Covering Theorem.

Borel-Covering Lemma

To prove Heine-Borel theorem, we shall firstly prove Borel-Covering Lemma. Let E ⊆ Rn. Let r : E → (0, ∞) be any function. That is, for every y ∈ E, r(y) is a positive real number and so we can consider the open ball Br(y)(y) with center at y and radius r(y) > 0. Clearly {Br(y)(y)}y∈E is an open covering of E. So if we wish to prove Heine-Borel theorem, we shall prove that if E is closed and bounded, then {Br(y)(y)}y∈E has a finite subcover. As we shall see this will be sufficient to prove Heine-Borel Theorem.

Theorem

Borel-Covering Lemma. If E is a closed and bounded subset of Rn and r : E → (0, ∞) is any function, then there exists N ∈ Z+ and finitely many points y1, y2, . . . , yN such that

N

• j=1

Br(yj)(yj) ⊇ E.

Proof of Borel-Covering Lemma

Let E be a closed and bounded subset of Rn and let r : E → (0, ∞) be any function. Then {Br(y)(y)}y∈E is an open covering of E. By Lindel¨

• f Covering Theorem, this open covering
• f E has a countable subcover. So there exists a sequence {yj}∞

j=1

in E such that

∞

• j=1

Br(yj)(yj) ⊇ E. Suppose for the contrary that for every k ∈ Z+,

k

• j=1

Br(yj)(yj) E. Then for every k ∈ Z+, there exists an element xk ∈ E such that xk ∈

k

• j=1

Br(yj)(yj). Consider the sequence {xk}∞

k=1 in E.

Since E is bounded, the sequence {xk}∞

k=1 is also bounded and

thus by Bolzano-Weierstrass Theorem in Rn, it has a convergent subsequence {xkν}∞

ν=1. Let x = lim ν→∞ xkν.

Since E is closed, the limit x of the convergent sequence {xkν}∞

ν=1

in E must also be in E. Since x ∈ E and E ⊆

∞

• j=1

Br(yj)(yj), we have x ∈ Br(yj0)(yj0) for some positive integer j0. By the characterization of limits of sequences in terms of open sets, we must have that the open ball Br(yj0)(yj0) containing the limit x = lim

ν→∞ xkν must contain also xkν for all large ν. That is,

there exists N ∈ Z+ such that for all integers ν ≥ N, the point xkν ∈ Br(yj0)(yj0).

But this contradicts with xk ∈

k

• j=1

Br(yj)(yj) for all k ∈ Z+, because there exists ν ∈ Z+ such that ν ≥ N and kν > j0 and so we must have xkν ∈

kν

• j=1

Br(yj)(yj) which implies that xkν ∈ Br(yj0)(yj0). This contradiction shows that it is not possible to construct the sequence {xk}∞

k=1 in E as described above, that is, it is not true

that

k

• j=1

Br(yj)(yj) E for all k ∈ Z+. So there exists some k ∈ Z+ such that

k

• j=1

Br(yj)(yj) ⊇ E.

This ends the proof of Borel-Covering Lemma which is the main tool in the following proof of Heine-Borel theorem. The first proof of Heine-Borel theorem that we shall give is the proof in your textbook that uses (Lindel¨

• f Covering Theorem and)

Borel-Covering Lemma. We shall also outline two other proofs of Heine-Borel theorem. Heine-Borel Theorem: A subset of Rn is compact if and only if it is closed and bounded.

First Proof of Heine-Borel Theorem

Let E be a closed and bounded subset of Rn. Let {Vα}α∈A be an open covering of E, that is,

• α∈A

Vα ⊇ E and Vα is open in Rn for every α ∈ A. For each y ∈ E, there exists an αy ∈ A such that y ∈ Vαy. Since Vαy is an open set, there exists a positive real number r(y) such that the open ball Br(y)(y) is contained in Vαy: y ∈ Br(y)(y) ⊆ Vαy for every y ∈ E. So here we have a function r : E → (0, ∞) and we can now use Borel-Covering Lemma: There exists N ∈ Z+ and finitely many points y1, y2, . . . , yN such that

N

• j=1

Br(yj)(yj) ⊇ E.

Since Vαy ⊇ Br(y)(y) for every y ∈ E, we obtain

N

• j=1

Vαyj ⊇ E. So we showed that every open covering of the closed and bounded subset E ⊆ Rn has a finite subcover. By definition of compactness, this means that E is compact. This was the harder part of Heine-Borel theorem. Let us also prove the easier part that every compact subset of Rn is bounded and closed. Let E be a compact subset of Rn. Consider the open balls with center at the origin 0 and radius k, a positive integer. The collection {Bk(0)}k∈Z+ of all these open balls clearly covers Rn and so also E. Then the open covering {Bk(0)}k∈Z+ of the compact set E must have a finite subcover, that is, there exists some N ∈ Z+ and positive integers k1, k2, . . . , kN such that

N

• j=1

Bkj(0) ⊇ E. So we obtain Bm(0) ⊇ E for m = max({k1, k2, . . . , kN}). That means for every x ∈ E, we have x ∈ Bm(0), that is, x < m. So E is bounded.

To prove that the compact set E is closed in Rn, we will show that E = E. Remember that the closure of a subset E ⊆ Rn is the union of E and all of its cluster points: E = E ∪ { all cluster points of E}. So to prove that E is closed, we must show that all cluster points

• f E is in E. Remember that a point x ∈ Rn is a cluster point (or

an accumulation point or a limit point) of the set E if for every real number r > 0, the intersection Br(x) ∩ E contains a point

• ther than the point x, that is,

(Br(x) \ {x}) ∩ E = ∅ for all real numbers r > 0. Suppose for the contrary that there exists a cluster point x of E such that x ∈ E. Since x ∈ E, we have E ⊆ Rn \ {x}. Consider the collection {Rn \ B1/k(x) | k ∈ Z+}

• f open sets. This is an open covering of Rn \ {x} and so an open

covering of the compact set E ⊆ Rn \ {x}. Since E is compact, this open covering of E must have a finite subcover.

Thus there exists N ∈ Z+ and positive integers k1, k2, . . . , kN such that

N

• j=1

Rn \ B1/kj(x) ⊇ E. Then we obtain Rn \ B1/m(x) ⊇ E for m = max({k1, k2, . . . , kN}). That implies E ∩ B1/m(x) = ∅. But since x is a cluster point of E, we have (Br(x) \ {x}) ∩ E = ∅ for every real number r > 0. In particular,

• B1/m(x) \ {x}
• ∩ E = ∅. Since B1/m(x) ⊆ B1/m(x), we
• btain the desired contradiction. This contradiction shows that

every cluster point of E must be in E. Thus the compact set E must be closed. This ends the first proof of Heine-Borel Theorem: A subset of Rn is compact if and only if it is closed and bounded.

Second Proof of Heine-Borel Theorem starts by proving that [a, b] ⊆ R is compact

Using the completeness of the real number system R, the following result is proved. See my handwritten notes for the topology of R for a proof of it:

Theorem

Every bounded closed interval in R is compact, that is, for all real numbers a ≤ b in R, the bounded closed interval [a, b] is compact. This is indeed the main step to prove Heine-Borel theorem in Rn by following the below steps.

Follow the below steps to give a second proof of the harder part of Heine-Borel Theorem, that is, to prove that a closed and bounded subset of Rn is compact:

◮ The cartesian product of two compact sets is compact:

If A is a compact subset of Rn and B is a compact subset of Rm, then their Cartesian product A × B is a compact subset

• f Rn+m.

◮ Cartesian product of compact sets is compact: If Ai is a

compact subset of Rni for each i = 1, 2, . . . , r where r ∈ Z+ and n1, n2, · · · , nr are positive integers, then their cartesian product A1 × A2 × · · · × Ar is a compact subset of Rn where n = n1 + n2 + . . . + nr.

◮ n-dimensional rectangles are compact: The n-dimensional

rectangle [a1, b1] × [a2, b2] × · · · × [an, bn] ⊆ Rn is compact, where for each k = 1, 2, . . . , n, [ak, bk] is a bounded closed interval in R (that is ak ≤ bk are finite real numbers).

◮ The last step is to prove that a closed and bounded subset E

• f Rn is compact. This is because since E is bounded, it is

contained in a large enough n-dimensional rectangle B. By the previous part, the n-dimensional rectangle B is compact. Since E ⊆ B, B is compact and E is closed, we obtain that E is compact, too. In this second proof of Heine-Borel theorem, as you have observed, the main step after you have proved that every bounded closed interval in R is compact, is the first claim: The cartesian product

• f two compact sets is compact. If x ∈ Rn and B is a compact

subset of Rm, then {x} × B is also compact. Indeed, more is true: If {Vα}α∈A is an open cover of {x} × B, then not only we have a finite subcover of {x} × B but also a finite subcover of U × B for some open set U ⊆ Rn containing x:

Theorem

If x ∈ Rn and B is a compact subset of Rm, and if {Vα}α∈A is an

• pen cover of {x} × B, then there exists an open subset U of Rn

containing x such that U × B is covered by a finite number of sets in the collection {Vα}α∈A.

Third Proof of Heine-Borel Theorem by Bisection Method

As explained for the last step of the second proof of Heine-Borel Theorem, it suffices to prove that n-dimensional rectangles are compact subsets of Rn. This can also be proved by the Bisection

• Method. Remember that the intersection of a nested sequence

{Ik}∞

k=1 of nonempty bounded closed intervals in R is nonempty.

We have used this to prove Bolzano-Weierstrass Theorem for sequences of real numbers. Similarly, prove that we have:

Theorem

If {Bk}∞

k=1 is a nested sequence of nonempty n-dimensional

rectangles, that is, B1 ⊇ B2 ⊇ · · · ⊇ Bk ⊇ Bk+1 ⊇ · · · and for each k ∈ Z+, Bk = Ik,1 × Ik,2 × · · · × Ik,n is an n-dimensional rectangle in Rn where Ik,1, Ik,2, . . . , Ik,n are nonempty bounded closed intervals in R, then

∞

• k=1

Bk = ∅. Now we use this theorem for the third proof of Heine-Borel

Let B = I1 × I2 × · · · × In be an n-dimensional rectangle where Ik = [ak, bk] is a bounded closed interval in R where ak ≤ bk are finite real numbers for each k = 1, 2, . . . , n. If δ is the length of the “diagonal” of this n-dimensional rectangle B, that is, δ =

• (b1 − a1)2 + (b2 − a2)2 + · · · + (bn − an)2,

then x − y ≤ δ for all x, y ∈ B. Suppose for the contrary that B is not compact. So there exists an

• pen cover {Vα}α∈A of B such that {Vα}α∈A has no finite

subcover. For each k = 1, 2, . . . , n, let ck = ak + bk 2 be the midpoint of the interval Ik = [ak, bk] that divides Ik into two subintervals I (1)

k

= [ak, ck] and I (2)

k

= [ck, bk]. Using these intervals, divide the n-dimensional rectangle B into n-dimensional rectangles of the form I (t1)

1

× I (t2)

2

× · · · × I (tn)

n

, where t1, t2, . . . , tn ∈ {1, 2}.

Clearly there are 2n such n-dimensional rectangles and their union is B. At least one of these n-dimensional rectangles, call it B1, cannot be covered by any finite subcollection of the open cover {Vα}α∈A of B (because otherwise the open cover {Vα}α∈A will have a finite subcover of B). We next subdivide the n-dimensional rectangle B1 to find an n-dimensional rectangle B2 that cannot be covered by any finite subcollection of the open cover {Vα}α∈A of B. Continuing the process we obtain a sequence {Bk}∞

k=1 of nonempty

n-dimensional rectangles contained in B such that for all k ∈ Z+,

◮ B1 ⊇ B2 ⊇ · · · ⊇ Bk ⊇ Bk+1 ⊇ · · · , ◮ Bk is not covered by any finite subcollection of the open cover

{Vα}α∈A of B.

◮ For all x, y ∈ Bk, x − y ≤ δ

2k .

By the previous theorem,

∞

• k=1

Bk = ∅. So there exists a point a ∈

∞

• k=1

Bk, that is, there exists a point a such that a ∈ Bk for all k ∈ Z+. Since {Vα}α∈A is an open cover of B, there exists α0 ∈ A such that Vα0 contains the point x. Since Vα0 is an open set, there exists a real number r > 0 such that Vα0 contains the open ball Br(a) with center at a and radius r. Choose k0 ∈ Z+ large enough such that δ 2k0 < r. By the properties of the sequence {Bk}∞

k=1 of

n-dimensional rectangles, for all x, y ∈ Bk0, x − y ≤ δ 2k0 . Since a ∈ Bk0, we obtain that for all x ∈ Bk0, x − a ≤ δ 2k0 < r. Thus Bk0 ⊆ Br(a) ⊆ Vα0. So the open set Vα0 contains the whole of

• Bk0. But this contradicts with the properties of the sequence

{Bk}∞

k=1 of n-dimensional rectangles; we have chosen each Bk

such that Bk is not covered by any finite subcollection of the open cover {Vα}α∈A of B. This contradiction ends our third proof of Heine-Borel theorem.

Distance between sets

The distance between two nonempty subsets A and B of Rn is defined to be the nonnegative real number d = inf({a − b | a ∈ A and b ∈ B}). So d ≥ 0 and for all a ∈ A and b ∈ B, a − b ≥ d. Of course d = 0 if A ∩ B = ∅. But even if A ∩ B = ∅, it may happen that d = 0.

The distance between a closed set and a compact set is positive if they are disjoint

Remember that we have previously observed that if A is a nonempty closed subset of Rn and B = {b} has only one element b ∈ Rn \ A, then the distance between A and the point b outside the set A is positive. More generally, prove the following:

Theorem

If A is a nonempty closed subset of Rn and B is a nonempty compact subset of Rn such that A ∩ B = ∅, then the distance between them is positive and so there exists a real number δ > 0 such that a − b ≥ δ for all a ∈ A and b ∈ B. Here it was necessary to assume that B is compact; give an example of two nonempty closed subsets A and B of Rn such that A ∩ B = ∅ and the distance between A and B is 0.

Open sets containing a compact set

The following result will be a good challenge for you to prove; use the previous result that the distance between a closed set and a compact set is positive. In the notation of the following theorem, the distance between the closed set Rn \ U and the compact set H is positive (Rn \ U and H are disjoint, that is, (Rn \ U) ∩ H = ∅ since H ⊆ U).

Theorem

Let H be compact subset of Rn. Let U be an open set in Rn that contains H: H ⊆ U. Then there exists a compact subset C of Rn such that the interior

• f C contains H and C is contained in U:

H ⊆ C o and C ⊆ U.

Definition of connected subsets of Rn

Let E ⊆ Rn.

◮ A pair of sets U and V are said to separate E if

• 1. E = U ∪ V ,
• 2. U ∩ V = ∅,
• 3. U and V are relatively open in E, and
• 4. U = ∅ and V = ∅.

◮ E is said to be CONNECTED if E cannot be separated by

any pair of sets. So if E is not connected, then there exists a pair of sets U and V that are relatively open in E such that E = U ∪ V , U ∩ V = ∅, U = ∅, V = ∅. If E is connected, then there exists no such pair of sets U and V .

Characterization of connectedness of E ⊆ Rn in terms of

• pen sets in Rn

Theorem

Let E ⊆ Rn.

◮ E is not connected if there exists a pair of open sets A and

B in Rn such that

• 1. E ⊆ A ∪ B,
• 2. A ∩ B = ∅,
• 3. A and B are open subsets of Rn, and
• 4. E ∩ A = ∅ and E ∩ B = ∅.

◮ If E is not connected, then there exists a pair of sets A and

B that satisfies the above four conditions. The first part is obvious; if we take U = E ∩ A and V = E ∩ B, then U and V will be relatively open sets in E that separate E and so E will not be connected. The proof of the second part (that is the converse of the first part) is not obvious. See your textbook for the proof of it.

Characterization of connectedness of E ⊆ Rn in terms of continuous functions from E to the two-element set {0, 1}

Theorem

A subset E of Rn is connected if and only if every continuous function f : E → {0, 1} is a constant function (where {0, 1} is the subset of R consisting of only the two elements 0 and 1). Proof: Assume that E is connected and f : E → {0, 1} is a continuous function. By the characterization of continuity in terms

• f relatively open sets, we have that the inverse image of every

relatively open subset of {0, 1} is relatively open in E. So the subsets U = f −1((−∞, 1/2) ∩ {0, 1}) = f −1({0}) and V = f −1((1/2, ∞) ∩ {0, 1}) = f −1({1}) of E are relatively open in

• E. Clearly we also have U ∩ V = f −1({0}) ∩ f −1({1}) = ∅. Since

f (E) ⊆ {0, 1}, we have U ∪ V = E. Since E is connected, at least

• ne of the sets U and V must be empty because otherwise U and

V separate E. If U = ∅, then E = V = f −1({1}), that is, f (x) = 1 for every x ∈ E which implies that f : E → {0, 1} is a constant

• function. Similarly, if V = ∅, then f is again a constant function.

Conversely assume that E is not connected. Then there exists a pair of sets U and V that are relatively open in E such that E = U ∪ V , U ∩ V = ∅, U = ∅, V = ∅. Consider the function f : E → R defined by f (x) = 0, if x ∈ U; 1, if x ∈ V . Since U = ∅ and V = ∅, f is not a constant function; it takes the value 0 at points in U = ∅ and it takes the value 1 at points in V = ∅. By using the characterization of continuity in terms of relatively open sets, we obtain that f : E → {0, 1} is a continuous function because for every open set W in R, the inverse image f −1(W ∩ {0, 1}) of the relatively open set W ∩ {0, 1} in {0, 1} is relatively open in E since f −1(W ∩ {0, 1}) =    U, if 0 ∈ W and 1 ∈ W ; V , if 1 ∈ W and 0 ∈ W ; U ∪ V = E, if 0 ∈ W and 1 ∈ W .

Closure of a connected subset of Rn is also connected

Using the above characterization of connectedness of a subset of Rn in terms of continuous functions from this set to {0, 1}, and also using the sequential characterization of continuity, and also the description of the closure E of a subset E of Rn in terms of limits of convergent sequences in E, prove the following:

Theorem

If a subset E of Rn is connected, then its closure E is also connected. Moreover, if E is connected and E ⊆ Y ⊆ E, then Y is also connected.

Union of connected subsets of Rn with nonempty intersection is connected

Theorem

If {Eα}α∈A is a collection of connected subsets of Rn such that the intersection

• α∈A

Eα is nonempty, then the union

• α∈A

Eα is also connected. Proof: Let T =

• α∈A

Eα and U =

• α∈A

Eα. Since T = ∅ by hypothesis, there exists an element a ∈ T, that is, a ∈ Eα for every α ∈ A. By the characterization of connected subsets of Rn, it suffices to show that every continuous function f : U → {0, 1} is

• constant. For such a continuous function f , its restriction

f|Eα : Eα → {0, 1} to Eα is also continuous and so must be a constant function since Eα is connected for every α ∈ A. Thus for every x ∈ Eα, f|Eα(x) = f|Eα(a) since a ∈ Eα for every α ∈ A. Hence f (x) = f (a) for every x ∈ U =

α∈A Eα, that is, f takes the

constant value f (a) on the set U. So the function f : U → {0, 1} is constant.

Connected components of a subset E of Rn

Let E ⊆ Rn. Let a ∈ E. Since the set {a} is connected, a belongs to at least

• ne connected subset of E. Consider the collection of all

connected subsets of E that contains a, say this collection is {Eα | α ∈ A}. By the previous theorem, since the intersection

• α∈A

Eα is nonempty (because it contains the point a), we obtain that the union

• α∈A

Eα is also connected. We call this union the connected component (or shortly component) of the set E that contains the point a. Denote this connected component by U(a). Then U(a) is the maximal connected subset of E that contains a, that is, it is a maximal element of the collection {Eα | α ∈ A}

• rdered by the subset relation.

Moreover, the collection {U(a) | a ∈ E} forms a partition of E. Be careful, for some points a, b ∈ E, it is of course possible that U(a) = U(b).

Properties of connected components of a subset E of Rn

Indeed, you can use the relation ∼ on E defined as follows: For each a, b ∈ E, a ∼ b if a and b belong to a connected subset of E. Then you prove that this relation ∼ is an equivalence relation on E and the equivalence classes of this relation give the above partition

• f E. It is a good exercise for you to prove the following:

Theorem

Let E ⊆ Rn.

◮ Connected components of E are connected. ◮ Connected components of E are relatively closed in E. ◮ Each connected subset of E is contained in a connected

component of E.

◮ Connected components of E are maximal connected subsets

• f E, that is, they are the maximal elements of the collection
• f all connected subsets of E ordered by the subset relation.

◮ Connected components of E forms a partition of E; that is,

any two distinct connected components of E are disjoint and the union of all connected components of E is E.

Connected subsets of R are just intervals

The intervals in R are the only connected subsets of R. Firstly prove the following property that characterizes intervals in R: For a subset E of R,

◮ E is an interval if and only if for all real numbers x and y,

x ∈ E, y ∈ E and x < y = ⇒ the interval [x, y] ⊆ E. Remember also that by an interval in R we mean any of the following sets in R where a ≤ b in R are finite real numbers: Bounded intervals: [a, b], (a, b), [a, b), (a, b], Unbounded intervals: [a, ∞), (a, ∞), (−∞, b), (−∞, b], (−∞, ∞). Note also that the empty set ∅ = (a, a) and singletons {a} = [a, a] are also intervals according to this definition of intervals.

Theorem

A subset E of the real numbers R is connected if and only if E is an interval.

Proof: Equivalently, we will prove that a subset E of the real numbers R is not connected if and only if E is not an interval. If E is not an interval, then by the property characterizing intervals there exist real numbers a, b, c such that a < c < b, a ∈ E and b ∈ E but c ∈ E. Then the open intervals A = (−∞, c) and B = (c, ∞) are open subsets of R such that a ∈ E ∩ A and b ∈ E ∩ B. So E ∩ A = ∅ and E ∩ B = ∅. And we also have E ⊆ A ∪ B and A ∩ B = ∅. Thus we have found open sets A and B in R such that E ⊆ A ∪ B, A ∩ B = ∅, E ∩ A = ∅, E ∩ B = ∅. By the characterization of connected subsets of R in terms of open subsets of R, this implies that E is not connected. Conversely assume that E is not connected. Then there exists a pair of sets U and V that are relatively open in E such that E = U ∪ V , U ∩ V = ∅, U = ∅, V = ∅. Firstly we show that U ∩ V = ∅ and U ∩ V = ∅.

Since U and V are relatively open sets in E, there exist open sets A and B in R such that U = E ∩ A and V = E ∩ B. Then ∅ = U ∩ V = U ∩ (E ∩ B) = (U ∩ E) ∩ B = U ∩ B and so U ⊆ R \ B. Since B is an open subset of R, the set R \ B is closed in R. Since U is contained in the closed set R \ B, we have U ⊆ R \ B. Hence U ∩ B = ∅ and so U ∩ V = ∅ since V ⊆ B. Similarly, it is shown that U ∩ V = ∅ by proving that V ⊆ R \ A. Since U = ∅ and V = ∅, there exist elements a ∈ U and b ∈ V . Let c = sup(U ∩ [a, b]). Then a ≤ c ≤ b. Since c = sup(U ∩ [a, b]), there exists a sequence in U ∩ [a, b] with limit c. So c is in the closure of the set U ∩ [a, b] and then c ∈ U ∩ [a, b] ⊆ U. Since U ∩ V = ∅, we must have c ∈ V . This then gives a ≤ c < b because we cannot have c = b since b ∈ V but c ∈ V . If we also have that c ∈ U, then we obtain c ∈ U ∪ V = E and so a < c < b with a, b ∈ E but c ∈ E which implies that E is not an interval.

Assume c ∈ U. Then c ∈ V since U ∩ V = ∅. If it were true that the open interval (c, b) is contained in V , then we would have that c is in V . So there exists a point c1 in the open interval (c, b) such that c1 ∈ V : c < c1 < b and c1 ∈ V . Since c = sup(U ∩ [a, b]) and a ≤ c < c1 < b, we cannot have that c1 ∈ U. Thus c1 ∈ U ∪ V = E. So we have a < c1 < b, a, b ∈ E but c1 ∈ E. This implies that E is not an interval. So in any case, we showed that E is not an interval if E ⊆ R is not connected. This ends the proof.

Image of a connected set under a continuous function is connected

Theorem

Let E ⊆ Rn and f : E → Rm be a function. Let H ⊆ E. If H is a connected set in Rn and f : E → Rm is a continuous function, then f (H) is a connected set in Rm. We shall later give the proof of this theorem.

Path connected subsets of Rn

A subset E of Rn is said to be path connected (or arcwise connected or pathwise connected) if for every pair of points a, b ∈ E, there exists a continuous function f : [0, 1] → E from the bounded closed interval [0, 1] to E such that f (0) = a and f (1) = b. So in a path connected space E, every pair of points is connected by a “path” or “curve” that lies wholly in E because we have said that the continuous function f is from the bounded closed interval [0, 1] to E. So f (t) ∈ E for every t ∈ [0, 1]. For a set E ⊆ Rn and points a, b ∈ E, a path in E is a continuous function f : [0, 1] → E from the bounded closed interval [0, 1] to E such that f (0) = a and f (1) = b. In this definition, we can take any closed interval [c, d] where c < d in R instead of the interval [0, 1]. Thus E is path connected means that every pair of points in E is joined by a path in E.

Convex subsets and star-shaped subsets of Rn

A subset E of Rn is said to be convex if for every pair of points a, b in E, the line segment L(a; b) = {(1 − t)a + tb | t ∈ [0, 1]} from a to b is contained in E. So clearly convex subsets of Rn are path connected. A subset E of Rn is said to be star-shaped if there exists a point a ∈ E such that for every point b in E, the line segment L(a; b) = {(1 − t)a + tb | t ∈ [0, 1]} from a to b is contained in E. Clearly a star-shaped set E is path connected because for any pair of points b1, b2 in E, the path from b1 to b2 in E is obtained by going through the line segment from b1 to a and then from a to b2.

Image of a path connected set under a continuous function is path connected

Since composition of continuous function is continuous, the following result is easily obtained:

Theorem

Let E ⊆ Rn and f : E → Rm be a function. Let H ⊆ E. If H is a path connected set in Rn and f : E → Rm is a continuous function, then f (H) is a path connected set in Rm.

Path connected sets are connected

The image of every continuous function f : I → Rn from an interval I in R is connected since the intervals are connected subsets of R. Using this we shall prove that:

Theorem

Every path connected subset of Rn is connected. What about the converse? Is every connected subset of Rn path connected?

Proof of (path connected)= ⇒(connected): Let E be a path connected subset of Rn. By the characterization of connected subsets of Rn in terms of functions to {0, 1}, it suffices to show that every continuous function f : E → {0, 1} is constant. We shall show that if f : E → {0, 1} is a continuous function, then f (a) = f (b) for every pair of points a, b in E and that means that f : E → {0, 1} is a constant function, it takes the same value at every point in E. Since E is path connected, there exists a path in E from a to b, that is, there exists a continuous function g : [0, 1] → E such that g(0) = a and g(1) = b. Let U = g([0, 1]). Since the interval [0, 1] is a connected subset of R, its image U = g([0, 1]) under the continuous function g is also

• connected. Since f : E → {0, 1} is continuous, so is its restriction

f|U : U → {0, 1} to U = g([0, 1]). Since U = g([0, 1]) is connected, the continuous function f|U : U → {0, 1} must be constant. In particular, f|U(a) = f|U(b) since U = g([0, 1]) contains the points a and b. Thus f (a) = f (b) for every pair of points a, b in E.

Product of intervals is path connected

We have previously given the proof of the fact that connected subsets of R are just intervals. The harder part of that proof was to show that every connected subset of R is an interval. The easier part was to show that intervals are connected; this is also obvious now since intervals in R are path connected subsets of R and so they are connected. Indeed you can prove easily that a subset E of R is path connected if and only if E is an interval in R. Similarly we have that Cartesian product of intervals is path connected:

Theorem

If I1, I2, . . . , In are intervals in R, then their Cartesian product I1 × I2 × · · · × In is a convex subset of Rn, and so it is a path connected and connected subset of Rn.

A connected subset of R2 that is not path connected

It is not true that every connected subset of Rn is path connected. The following are examples of connected subsets of Rn that are connected but not path connected. Its rigorous proof is left to you as a challenging exercise.

◮ The subset E of R2 defined by

E = ({0} × [−1, 1]) ∪ {(x, sin(1/x)) | x ∈ (0, 1]}. is connected but not path connected.

◮ The subset F of R2 defined by

F = {(0, 1)} ∪ ([0, 1] × {0}) ∪ ({1/k | k ∈ Z+} × [0, 1]). is connected but not path connected.

Open and connected subsets of Rn are path connected

The answer to our question “Is every connected subset E of Rn path connected?” will be yes if we further assume that the set E is

• pen:

Theorem

Every open and connected subset of Rn is path connected. The main observation to prove this theorem is that an open ball is a convex set and so path connected; if we take any point b in an

• pen ball Br(a) where r > 0 is a real number and a ∈ Rn, then the

line segment L(a; b) = {(1 − t)a + tb | t ∈ [0, 1]} from a to b is contained in Br(a) (Why? Prove it.).

Proof of (open and connected)= ⇒(path connected): Let E be an open and connected subset of Rn. Fix a point a in E. If we take an arbitrary point b in E, then there is either a path in E from a to b or there exists no path in E from a to b. So we consider the following subsets U and V of E: U = {b ∈ E | there is a path in E from a to b} V = {b ∈ E | there exists no path in E from a to b} Then E = U ∪ V and U ∩ V = ∅. Here U = ∅ since a ∈ U clearly. If we can show that U and V are (relatively) open in the open set E, then U and V will separate E if V = ∅. Since E is connected, E has no separation. Thus we must have V = ∅ and so E = U. But E = U means that every point of E can be joined to a by a path in

• E. Since a ∈ E is also arbitrary, we obtain that every pair of points

in E can be joined by a path in E. So E is path connected. What remains is to show that U and V are relatively open in the open set E. Since E is an open subset of Rn, its relatively open subsets are open sets in Rn. So we will show that U and V are open.

Let b ∈ U. Then there is a path in E from a to b. Since E is an

• pen set, there exists a real number r > 0 such that E contains

the open ball Br(b). We shall show that Br(b) ⊆ U. Let c ∈ Br(b). Since the open balls are convex, the line segment from b to c lies in the open ball Br(b) and so we can join the point a to the point c by a path in E by using firstly the path from a to b in E and then the line segment from b to c. So the point c is also in

• U. This shows that U is an open subset of Rn.

Similarly, one can show that V is open: Because for a point b ∈ V , if a point c in an open ball Br(b) ⊆ E could be joined to a by a path in E, then adding the line segment from c to b to this path from a to c would give a path in E from a to b, which contradicts b ∈ V .

Open and connected subsets of Rn are polygonally connected

Let E be a subset of Rn. A path f : [0, 1] → E is said to be a polygonal path if the image f ([0, 1]) is the union of a finite number of line segments. The set E is said to be polygonally connected if every pair of points can be joined by a polygonal path in E. Thus E is polygonally connected if and only if for every pair of points a and b in E, there exist finitely many points x0, x1, x2, . . . , xN in E for some N ∈ Z+ such that x0 = a, xN = b and for each k = 1, 2, . . . , N, the line segment L(xk−1; xk) = {(1 − t)xk−1 + txk | t ∈ [0, 1]} from xk−1 to xk is contained in E. The proof of the previous theorem also works when we take polygonal paths instead of paths and this gives us:

Theorem

Every open and connected subset of Rn is polygonally connected.

Connected components of an open subset of Rn are open

Let E be an open subset of Rn. Let U be a connected component of E. Let a ∈ U. Since a ∈ U and U is a connected component of E, we must have that U = U(a) where U(a) is the union of all connected subsets of E containing a. Since E is an open set, there exists a real number r > 0 such that the open ball Br(a) is contained in E. Since open balls are connected, the open ball Br(a) containing a must be in the connected component U(a) because U(a) is the union of all connected subsets of E containing a. Thus for each a ∈ U, we found a real number r > 0 such that Br(a) ⊆ U. This means that U is open. Hence we showed the first part of the following theorem. The second part is obtained by Lindel¨

• f Covering Theorem since we already know that E is the

union of the collection of connected components of E that are pairwise disjoint.

Theorem

Let E be an open subset of Rn. Then:

◮ Every connected component of E is also open in E. ◮ There exists a countable collection of pairwise disjoint open

connected subsets of Rn whose union is the set E. Moreover, the expression of E in this form is unique.

Region, open region, closed region

According to the previous theorem, an open subset of Rn can be expressed as a (countable) union of open connected subsets of Rn. When working with functions, we sometimes restrict the domain of the functions to open connected subsets of Rn. Some textbooks use the following terminology, although not standard:

◮ A region A in Rn is an open connected subset U of Rn plus

some boundary points of U (may be none of ∂U or may be some part of ∂U or may be all of ∂U).

◮ An open region A in Rn is an open connected subset of Rn. In

some textbooks (especially in complex analysis), an open connected set U is said to be a domain.

◮ A closed region A in Rn is the closure of an open connected

subset U of Rn, that is, A = U = U ∪ ∂U.

Homotopic paths in a subset E of Rn

Let E be a subset of Rn. Let a, b ∈ E. Let f : [0, 1] → E and g : [0, 1] → E in E starting at the point a and ending at the point b: f (0) = g(0) = a and f (1) = g(1) = b. The paths f : [0, 1] → E and g : [0, 1] → E in E are said to be path homotopic if there exists a continuous function H : [0, 1] × [0, 1] → E such that

• 1. H(t, 0) = f (t) and H(t, 1) = g(t) for all t ∈ [0, 1], and
• 2. H(0, s) = a and H(1, s) = b for all s ∈ [0, 1].

The continuity of the function H : [0, 1] × [0, 1] means that we have a continuous way of deforming the path f to the path g via the following paths fs : [0, 1] → E, s ∈ [0, 1], in E that starts at the point a and ends at the point b: For each s ∈ [0, 1], the paths fs : [0, 1] → E defined by fs(t) = H(t, s), t ∈ [0, 1] are such that f0 = f , f1 = g, and for all s ∈ [0, 1], fs(0) = a and fs(1) = b.

Simply Connected Subsets of Rn: Subsets of Rn without “holes”

Let E be a subset of Rn. A path f : [0, 1] → E in E is said to be a closed path (or a loop) if it starts and ends at the same point, that is, f (0) = f (1). A path f : [0, 1] → E in E is said to be a constant path if it is a constant function, that is, for some a ∈ E, f (t) = a for all t ∈ [0, 1]. A subset E of Rn is said to be simply connected if it is path connected and every closed path in E is homotopic to a constant path; equivalently, E is path connected and every pair of paths in E are homotopic. In your elective courses in algebraic topology, you shall learn the fundamental group of a topological space which is the first homotopy group of this topological space. Being simply connected will mean that it is path connected and its fundamental group is trivial, that is, it consists of just one element (the identity element).

Jordan Curves: Simple closed paths

A path f : [0, 1] → Rn in Rn is said to be a simple path if it does not intersect itself except possibly at the endpoints, that is, for all s, t ∈ [0, 1], if 0 ≤ s < t < 1, then f (s) = f (t) (but it may be that f (0) = f (1)). A path f : [0, 1] → Rn in Rn is said to be a simple closed path if it is a closed path that does not intersect itself except the endpoints, that is, f (0) = f (1) and for all s, t ∈ [0, 1], if 0 ≤ s < t < 1, then f (s) = f (t). Simple closed paths in the two-dimensional plane R2 are called Jordan curves and we have the following important theorem whose proof is not easy for an arbitrary simple closed path in R2. You can find its proof in some textbooks on topology; for example, see the book Topology by James Munkres (2nd Edition, 2003, Prentice Hall).

Jordan Curve Theorem

Theorem

Jordan Curve Theorem. Let f : [0, 1] → R2 be a simple closed path in R2 and let C be the image f ([0, 1]) of the path f : [0, 1] → R2 in R2. Then the Jordan curve C separates the plane R2 into two disjoint open connected subsets B and U having C as their common boundary such that one of them is bounded, say B, and the other is unbounded, say U that is,

• 1. R2 \ C = B ∪ U and B ∩ U = ∅.
• 2. B and U are open connected subsets of R2.
• 3. B is a bounded subset of R2; it is called the interior (or the

inner region) of the Jordan curve C.

• 4. U is an unbounded subset of R2; it is called the exterior (or

the outer region) of the Jordan curve C.

• 5. ∂B = C and ∂U = C, that is, both of B and U have

boundary C.

Image of a compact set under a continuous function is compact

Theorem

Let E ⊆ Rn and f : E → Rm be a function. Let H ⊆ E. If H is a compact set in Rn and f : E → Rm is a continuous function, then f (H) is a compact set in Rm. Proof: Let {Vα}α∈A be an open covering of f (H) in Rm. Then {f −1(Vα)}α∈A is a covering of H because f (H) ⊆

• α∈A

Vα implies that H ⊆ f −1(f (H)) ⊆ f −1

α∈A

Vα

• =
• α∈A

f −1(Vα). Since f : E → Rm is continuous, for each α ∈ A, the inverse image f −1(Vα) of the open set Vα is relatively open in E and so for some

• pen set Uα in Rn

f −1(Vα) = Uα ∩ E.

Then {Uα}α∈A is an open covering of the compact set H and so must have a finite subcover, that is, for some finite subset B ⊆ A, {Uα}α∈B covers H and so {Vα}α∈B covers f (H). This is because the function f : E → Rm has domain E and so H ⊆

• α∈B

Uα implies that f (H) ⊆ f

• E ∩
• α∈B

Uα

• =
• α∈B

f (E∩Uα) =

• α∈B

f (f −1(Vα)) ⊆

• α∈B

Vα. Thus we showed that every open covering of f (H) has a finite subcover and that means that f (H) is a compact subset of Rm. This ends the proof.

One-to-one continuous functions on compact sets has a continuous inverse

Theorem

Let H ⊆ Rn and f : H → Rm be a function. Assume the function f : H → Rm is one-to-one and so we can consider the inverse function f −1 : f (H) → Rn. For each y ∈ f (H), f −1(y) = x where x is the unique element in H such that f (x) = y. If H is a compact subset of Rn and f : H → Rm is a one-to-one continuous function, then the inverse function f −1 : f (H) → Rn is continuous. To prove this, remember the following results:

◮ Closed subsets of compact sets are compact: If H is a

compact subset of Rn and A is a closed subset of Rn such that A ⊆ H, then A is also compact.

◮ A function g : E → Rn, where E ⊆ Rm, is continuous if and

• nly if for every closed subset K of Rm, the inverse image

g−1(K) is a relatively closed subset of E.

Proof: Let g = f −1 : f (H) → Rn. Let K be a closed subset of Rn. Since the image of the function g = f −1 : f (H) → Rn is H, we have g−1(K) = g−1(K ∩ H). Since H is a compact subset of Rn, H is closed. Since K and H are also closed, their intersection K ∩ H is also closed. Thus K ∩ H is a closed subset of Rn such that K ∩ H is contained in the compact subset H of Rn. Then K ∩ H is also compact. We have: g−1(K) = g−1(K ∩ H) = (f −1)−1(K ∩ H) = f (K ∩ H). Since f : H → Rm is continuous and K ∩ H is compact, f (K ∩ H) is compact, too. So f (K ∩ H) is a closed subset of Rm. Thus we showed that for every closed subset K of Rn, g−1(K) = f (K ∩ H) is a closed subset of Rm. Then since f (K ∩ H) ⊆ f (H), we obtain that for every closed subset K of Rn g−1(K) = f (K ∩ H) = f (K ∩ H) ∩ f (H) is relatively closed in f (H). That implies that the function f −1 = g : f (H) → Rn is continuous.

Continuity on compact sets implies uniform continuity

The proofs of the following results on uniform continuity are left to you as an exercise; they are obtained using the same ideas as in the case of real-valued functions of a real variable.

Theorem

Let E ⊆ Rn and f : E → Rm be a function. If E is a compact set and f : E → Rm is a continuous function, then f : E → Rm is uniformly continuous.

Theorem

Let D ⊂ E ⊆ Rn be such that D is dense in E, that is D = E. Let f : D → Rm be a function. If f : D → Rm is a uniformly continuous function on D, then f : D → Rm has a continuous extension to D = E, that is, there exists a continuous function g : E → Rm such that g(x) = f (x) for all x ∈ D.

Theorem

Let D be a bounded subset of Rn and let f : D → Rm be a function. Let E = D. Then:

◮ E is closed and bounded, and so compact by Heine-Borel

theorem.

◮ f : D → Rm is uniformly continuous on D if and only if

f : D → Rm has a continuous extension to D = E, that is, there exists a continuous function g : E → Rm such that g(x) = f (x) for all x ∈ D.

Extreme value theorem for continuous real-valued functions on a compact set

A real-valued continuous function on a compact set is bounded and attains its maximum and minimum value on its compact domain:

Theorem

Let H ⊆ Rn. Let f : H → R be a real-valued function. If H is compact and f is continuous on H, then we have:

◮ f is bounded on H, that is, the set {f (x) | x ∈ H} is a

bounded set of real numbers, and so

◮ M = sup x∈H

f (x) = sup({f (x) | x ∈ H}) is a finite real number, and

◮ m = inf x∈H f (x) = inf({f (x) | x ∈ H}) is a finite real number. ◮ Moreover, there exist xM and xm in H such that

M = f (xM) and m = f (xm).

Thus M is the maximum value and m is the minimum value of

• f the continuous function f : H → R on the compact set H:

◮ f (xM) = M = max x∈H f (x) = max({f (x) | x ∈ H}), ◮ f (xm) = m = min x∈H f (x) = min({f (x) | x ∈ H})

That is, the set f (H) = {f (x) | x ∈ H} has a maximum element and a minimum element if f : H → R is a continuous real-valued function on the compact set H. Proof: Since H is compact and f : H → R is continuous, the image f (H) = {f (x) | x ∈ H} is compact, too. Thus by Heine-Borel theorem, f (H) is closed and bounded. Since it is bounded, M = sup(f (H)) and m = inf(f (H)) are finite real

• numbers. Since M and m are the supremum and infimum of the

set f (H), there exist sequences {ak}∞

k=1 and {bk}∞ k=1 in the set

f (H) such that M = lim

k→∞ ak and m = lim k→∞ bk. Since the set f (H)

is closed, the limits of the convergent sequences {ak}∞

k=1 and

{bk}∞

k=1 in the set f (H) must be in the set f (H). That means

M ∈ f (H) and m ∈ f (H), that is, there exist xM and xm in H such that M = f (xM) and m = f (xm).

Indeed, this extreme value property characterizes compact subsets; the proof of this is left as an exercise:

Theorem

A nonempty subset H of Rn is compact if and only if for every continuous function f : H → R, there exist xM and xm in H such that f (xm) ≤ f (x) ≤ f (xM) for all x ∈ H.

Image of a connected set under a continuous function is connected

Theorem

Let E ⊆ Rn and f : E → Rm be a function. Let H ⊆ E. If H is a connected set in Rn and f : E → Rm is a continuous function, then f (H) is a connected set in Rm. Proof: Equivalently, we shall prove that if f : E → Rm is a continuous function, H ⊆ E ⊆ Rn and f (H) is not connected, then H is not connected. Since f (H) is not connected, there exists a pair of relatively open sets U and V in f (H) that separate f (H): f (H) = U ∪ V , U ∩ V = ∅, U = ∅, V = ∅. Let A = f −1(U) and B = f −1(V ). Our claim is that A and B are relatively open sets in H that separate H and so H will not be connected.

Let g : H → f (H) be the restriction of the continuous function f : E → Rm, that is, for all x ∈ E, we define g(x) = f (x). Since f is a continuous function, so is its restriction g : H → f (H). Now we can use the characterization of continuity in terms of relatively

• pen sets for the continuous function g : H → f (H) to obtain that

A = f −1(U) = g−1(U) and B = f −1(V ) = g−1(V ) are relatively

• pen sets in H, and you can show easily that they separate H:

H = A ∪ B, A ∩ B = ∅, A = ∅, B = ∅. That means H is not connected and ends our proof.

Intermediate value theorem for real-valued continuous functions on connected sets

Theorem

Let E ⊆ Rn and f : E → R be a real-valued function. If E is connected and f : E → R is a continuous function, then f takes every value between any two of its values, that is, if y ∈ R is between f (a) and f (b) for some a, b ∈ E, then there exists x ∈ E such that f (x) = y. The reason for this is that f (E) is a connected subset of R and so it is an interval. Proof: Since E is connected, the image f (E) must be a connected subset of R. But connected subsets of R are just intervals. So f (E) is an interval. If y ∈ R is between f (a) and f (b) for some a, b ∈ E, then since f (a) and f (b) are in the interval f (E), the real number y between f (a) and f (b) is also in the interval f (E). Hence f (x) = y for some x ∈ E.

Indeed, this intermediate value property characterizes connected subsets because remember that a subset I of R is an interval if and

• nly if for all real numbers x and y,

x ∈ I, y ∈ I and x < y = ⇒ the interval [x, y] ⊆ I.

Theorem

Let E ⊆ Rn. The subset E ⊆ Rn is connected if and only if for every continuous function f : E → R, f (E) is an interval in R.

The n-dimensional rectangle [a1, b1] × [a2, b2] × · · · × [an, bn] ⊆ Rn is compact and connected

Theorem

Let a1, a2, . . . , an, b1, b2, . . . , bn be real numbers such that aj ≤ bj for each j = 1, 2, . . . , n. Then we have n bounded closed intervals [a1, b1], [a2, b2], . . . , [an, bn] whose cartesian product gives the n-dimensional rectangle B in Rn: B = [a1, b1] × [a2, b2] × · · · × [an, bn] = {(x1, x2, . . . , xn) ∈ Rn | aj ≤ xj ≤ bj for all j = 1, 2, . . . , n} ⊆ Rn This n-dimensional rectangle B is connected and compact. The n-dimensional rectangle B is connected since it is path-connected; it is even a convex subset of Rn. By Heine-Borel theorem, since the n-dimensional rectangle B is closed and bounded, it is compact.

Graph of a continuous real-valued function of a real variable

Theorem

Let I be a closed interval (may be unbounded) in R and let f : I → R be a function. Then f : I → R is continuous if and only if the graph of f , that is, the set {(x, f (x)) | x ∈ I}, is a closed and connected subset in R2. This is the rough idea that the graph of a continuous function can be drawn on a paper without leaving the pencil from the paper. See your textbook for the proof of this theorem. By Heine-Borel theorem, compact subsets of R are just the bounded closed subsets

• f R. Since continuous functions preserve compactness, we obtain:

Theorem

Let I be a bounded closed interval (so compact) in R and let f : I → R be a function. Then f : I → R is continuous if and only if the graph of f , that is, the set {(x, f (x)) | x ∈ I}, is a bounded, closed and connected subset in R2.

The image of a bounded closed interval of real numbers under a continuous function is again a bounded closed interval

You have already seen the proof of the following theorem but now you can prove it using compactness and connectedness.

Theorem

Let f : [a, b] → R be a continuous function where [a, b] is a bounded closed interval in R, that is, a ≤ b in R. Then:

◮ Extreme value theorem: M = sup x∈[a,b]

f (x) and m = inf

x∈[a,b] f (x) are finite real numbers and there exist xM and

xm in [a, b] such that f (xm) = m and f (xM) = M.

◮ Intermediate value theorem: If y is a real number such that

y is between f (c) and f (d) for some c, d ∈ [a, b], then there exists a real number x between c and d such that f (x) = y.

◮ The image f ([a, b]) is again a bounded closed interval [m, M]

where M = sup

x∈[a,b]

f (x) and m = inf

x∈[a,b] f (x).

Monotone sequence of functions on a compact set

A sequence {fk}∞

k=1 of real-valued functions defined on a subset H

• f Rn is said to be:

◮ pointwise increasing on H if fk(x) ≤ fk+1(x) for all x ∈ H; ◮ pointwise decreasing on H if fk(x) ≥ fk+1(x) for all x ∈ H; ◮ pointwise monotone on H if it is pointwise increasing or

pointwise decreasing on H.

Theorem

Dini’s Theorem for Pointwise Convergent Monotone Sequences of Functions on a Compact Set. Suppose that {fk}∞

k=1 is a pointwise monotone sequence of

real-valued continuous functions defined on a compact subset H

• f Rn. If fk → f pointwise on the compact set H as k → ∞ (that

is, limk→∞ fk(x) = f (x) for all x ∈ H) for some real-valued function f defined on H, then fk → f uniformly on the compact set H, that is, for every ǫ > 0, there exists N ∈ Z+ such that for all integers k ∈ Z+, k ≥ N = ⇒ for all x ∈ H, |fk(x) − f (x)| < ǫ.

Limit of a monotone sequence of real-valued functions on [a, b]

See your textbook for the proof of Dini’s theorem. In the case of real-valued continuous functions on a bounded closed interval, we know that uniform convergence enables us to make the following change with the limit and integral:

Theorem

Suppose that {fk}∞

k=1 is a pointwise monotone sequence of

real-valued continuous functions defined on a bounded closed interval [a, b] in R. If fk → f pointwise on the bounded closed interval [a, b] as k → ∞ (that is, limk→∞ fk(x) = f (x) for all x ∈ [a, b]) for some real-valued function f defined on [a, b], then fk → f uniformly on the bounded closed interval [a, b] as k → ∞, and so f is continuous on [a, b]. Moreover, we have: lim

k→∞

b

a

fk(t) dt = b

a

• lim

k→∞ fk(t)

• dt =

b

a

f (t) dt.

Sets of measure zero

We have seen that a continuous real-valued function on a bounded closed interval [a, b] in R is Riemann integrable on [a, b]. The important theoretical question is to characterize integrable real-valued functions on a bounded closed interval [a, b]. How far are they from being continuous? We shall see the Lebesgue’s criterion for Riemann integrability: It says that a bounded function f : [a, b] → R is Riemann integrable on [a, b] if the set of discontinuities of f has “measure zero”, that is, it has a small size in the sense defined as follows: A subset E of R is said to have measure zero if for every ǫ > 0, there is a countable collection {Ik}∞

k=1 of intervals that covers E

such that

∞

• k=1

|Ik| < ǫ, that is, the sum of the lengths of all of the intervals in the covering {Ik}∞

k=1 of E is less than ǫ. Here for an interval I in R, we denote

its length by |I|.

The integrable functions are the bounded almost everywhere continuous functions

A function f : [a, b] → R, where [a, b] is a bounded closed interval in R, is said to be almost everywhere continuous on [a, b] if the set of discontinuities of f on [a, b], that is, the set {x ∈ [a, b] | f is discontinuous at x} is a set of measure zero.

Theorem

Lebesgue’s Criterion for Riemann Integrability. Let f : [a, b] → R be a bounded function. Then f is Riemann integrable on [a, b] if and only if f is almost everwhere continuous

• n [a, b].

In particular, since countable subsets of R have measure zero, we have that if f : [a, b] → R is a bounded function and has only countably many points of discontinuity on [a, b], then f is Riemann integrable on [a, b].

Measure theory and Lebesgue Integral

Your textbook gives the proof of the above theorem; you need to use the notion of oscillation of a function at a point. I leave the proof of it to your elective courses on real analysis where Measure Theory and Lebesgue Integral are studied. There is a better integral theory, called Lebesgue integral. We have above defined sets of measure zero but it is possible to give a “measure” (“length”) for a larger collection of subsets of R. Measure theory and Lebesgue integral is an important subject for any work on real analysis and I advice you to study them by taking related elective courses in your next years.

Functions f : Rn → R with compact support

Let f : Rn → R be a function.

◮ The support of f is the closure of the set of points at which f

is nonzero; it is denoted by spt(f ) (or by support(f )): spt(f ) = {x ∈ Rn | f (x) = 0}. So for every a ∈ Rn, if a is not in spt(f ), then there exists an

• pen subset U of Rn that contains a such that f (x) = 0 for

all x ∈ U.

◮ The function f : Rn → R is said to have compact support if

its support is a compact subset of Rn. Since by definition, spt(f ) is the closure of a subset of Rn, it is a closed set. So by Heine-Borel Theorem, a function f : Rn → R has a compact support if and only if spt(f ) is a bounded subset of Rn, or equivalently the set {x ∈ Rn | f (x) = 0} is a bounded subset of Rn.

Sum and product of functions with compact support have compact support

Theorem

If functions f : Rn → R and g : Rn → R have compact support, and α ∈ R then:

◮ Their sum f + g : Rn → R also has compact support and

spt(f + g) ⊆ spt(f ) ∪ spt(g).

◮ Their product fg : Rn → R also has compact support and

spt(fg) ⊆ spt(f ) ∩ spt(g).

◮ αf : Rn → R also has compact support and

spt(αf ) = spt(f ) if α = 0.

Functions with compact support and continuous partial derivatives up to order p: Cp

c (Rn)

Let p ∈ Z+. Let N ∈ Z+. If fk : Rn → R is a function with compact support for k = 1, 2, . . . , N, then their sum

N

• k=1

fk : Rn → R is also a function with compact support. Moreover, if each fk has continuous partial derivatives of all orders up to p, then their sum

N

• k=1

fk : Rn → R also has continuous partial derivatives of all orders up to p. We denote by Cp

c (Rn) the collection of all functions from Rn to R

that have compact support and that have continuous partial derivatives of all orders up to p: That is, a function f : Rn → R is in Cp

c (Rn) if and only if f has compact support and f has

continuous partial derivatives of all orders up to p. As noted above, if f1, f2, . . . , fN are in Cp

c (Rn), then their sum N

• fk : Rn → R is also in Cp

c (Rn).

Functions with compact support and partial derivatives of all orders: C∞

c (Rn)

We denote by C∞

c (Rn) the collection of all functions from Rn to R

that have compact support and that have continuous partial derivatives of all orders: That is, a function f : Rn → R is in C∞

c (Rn) if and only if f has compact support and f has continuous

partial derivatives of all orders. So a function f : Rn → R is in C∞

c (Rn) if and only if f is in

Cp

c (Rn) for all p ∈ Z+:

C∞

c (Rn) = ∞

• p=1

Cp

c (Rn).

If N ∈ Z+ and f1, f2, . . . , fN are in C∞

c (Rn), then their sum N

• k=1

fk : Rn → R is also in C∞

c (Rn).

The function e−1/x2

Theorem

The function f : R → R defined for all x ∈ R by f (x) =

• e−1/x2,

if x = 0; 0, if x = 0. has derivatives of all orders on R and f (k)(0) = 0 for all k ∈ Z+. You shall remember this function as an example of a function in C∞(R), that is, f is a function with (continuous) derivatives of all

• rders on R, but it is not analytic because the Taylor series of f

with center 0 is identically zero since f (k)(0) = 0 for all k ∈ Z+:

∞

• k=1

f (k)(0) k! xk =

∞

• k=1

k!xk = 0 for all x ∈ R. But of course f (x) = e−1/x2 > 0 for all real numbers x = 0.

There is no analytic function in C∞

c (R) except the

identically zero function

We shall use the function f : R → R defined for all x ∈ R by f (x) =

• e−1/x2,

if x = 0; 0, if x = 0. to give a function in C∞

c (R) that is not the constant zero function.

It is no surprise to use this function that is not analytic because an analytic function that has compact support must be the identically zero function; this is an exercise for you to review analytic functions and the property analytic continuation needed to prove it.

A function in C∞

c (R) that is positive in an open interval

(a, b) and zero outside

Theorem

For every a < b, there is a function ϕ : R → R that has derivatives

• f all orders and that has compact support such that for all real

numbers t, we have

◮ t ∈ (a, b)

= ⇒ ϕ(t) > 0,

◮ t ∈ (a, b)

= ⇒ ϕ(t) = 0. This function ϕ is in C∞

c (R) and spt(ϕ) = [a, b].

We can define the function ϕ : R → R by using the function e−1/x2 as follows: For all t ∈ R, define ϕ(t) =

• e−1/(t−a)2e−1/(t−b)2,

if t ∈ (a, b); 0, if t ∈ (a, b).

A nonzero function in C∞(R) that is 0 to the left of (0, δ) and that is 1 to the right of (0, δ) and between 0 and 1 on the interval (0, δ)

Using the previous theorem, we know that for a given real number δ > 0, there exists a function ϕ in C∞

c (R) such that ◮ ϕ(t) > 0 for all real numbers t ∈ (0, δ), and ◮ ϕ(t) = 0 for all real numbers t ∈ (0, δ).

The function ψ : R → R defined for all t ∈ R by ψ(t) = t ϕ(u) du δ ϕ(u) du satisfies the properties in the following theorem:

Theorem

For each real number δ > 0, there is a function ψ : R → R that has derivatives of all orders such that

◮ 0 ≤ ψ(t) ≤ 1 for all t ∈ R, ◮ ψ(t) = 0 for t ≤ 0, and ◮ ψ(t) = 1 for t ≥ δ.

Thus the function ψ is in C∞(R), ψ is constant 0 on (−∞, 0] and constant 1 on [δ, ∞) and ψ([0, δ]) = [0, 1].

C∞ version of Urysohn’s Lemma (for normal topological spaces)

See your textbook for the proof of the following theorem that uses the functions ϕ ∈ C∞

c (R) and ψ ∈ C∞(R) defined above; this

result leads directly to partitions of unity.

Theorem

Let H be a compact nonempty subset of Rn. Let V be an open subset of Rn that contains H: H ⊆ V , H is compact and V is open. Then there exists a function h : Rn → R that has compact support and that has (continuous) partial derivatives of all orders (that is the function h is in C∞

c (Rn)) such that ◮ 0 ≤ h(x) ≤ 1 for all x ∈ Rn, ◮ h(x) = 1 for all x ∈ H, and ◮ spt(h) ⊂ V .

C∞ partitions of unity on Ω ⊆ Rn subordinate to an open covering {Vα}α∈A of Ω

Theorem

Let ∅ = Ω ⊆ Rn. Let {Vα}α∈A be an open covering of Ω. Then for each j ∈ Z+, there exists a function ϕj : Rn → R with compact support and an index αj ∈ A such that the sequence {ϕj}∞

j=1 of functions in C∞ c (Rn) and the sequence {αj}∞ j=1 of

indices in A satisfy the following properties:

◮ ϕj ≥ 0 for all j ∈ Z+. ◮ spt(ϕj) ⊂ Vαj for all j ∈ Z+. ◮ ∞

• j=1

ϕj(x) = 1 for all x ∈ Ω.

◮ If H is a nonempty compact subset of Ω, then there is a

nonempty open set W H and a positive integer N such that ϕj(x) = 0 for all integers j ≥ N and for all x ∈ W . In particular,

N

• j=1

ϕj(x) = 1 for all x ∈ W .

Cp partitions of unity on an open set Ω ⊆ Rn subordinate to an open covering

Let ∅ = Ω ⊆ Rn. Let {Vα}α∈A be a covering of Ω. A sequence {ϕj}∞

j=1 of functions from Rn to R is called a partition

• f unity on Ω subordinate to the covering {Vα}α∈A if Ω and all

the Vα’s are open and nonempty, the ϕj’s are all continuous with compact support and they satisfy the four properties in the above theorem for some sequence {αj}∞

j=1 of indices in A. If moreover for

p ∈ Z+ ∪ {0, ∞}, the functions ϕj are also Cp on Ω, that is, if they have continuous partial derivatives of all orders up to p on the

• pen set Ω, then this partition of unity on Ω is said to be a Cp

partition of unity on Ω subordinate to the covering {Vα}α∈A. By the above theorem, given any open covering {Vα}α∈A of any nonempty open set Ω ⊆ Rn, there exists a C∞ partition of unity on Ω subordinate to {Vα}α∈A, and so for every p ∈ Z+ ∪ {0, ∞}, this C∞ partition of unity is a Cp partition of unity on Ω subordinate to {Vα}α∈A.

Decomposition of a function into a series of functions with small support

Cp partitions of unity can be used to decompose a function f into a sum of functions fj that have small support and are as smooth as f . For example, let f : Ω → R be defined on a set Ω ⊆ Rn. Let p ∈ Z+ ∪ {0, ∞} and {ϕj}∞

j=1 be a Cp partition of unity on Ω

subordinate to a covering {Vj}∞

j=1 of Ω.

For each j ∈ Z+, let fj = f ϕj be the product of the functions f and ϕj, that is, fj(x) = f (x)ϕj(x). Then for all x ∈ Ω, f (x) = f (x)

∞

• j=1

ϕj(x) =

∞

• j=1

f (x)ϕj(x) =

∞

• j=1

fj(x). If f is C p on Ω, that is, if f has continuous partial derivatives of all

• rders up to p, then each fj = f ϕj is also C p on Ω and

spt(fj) ⊆ spt(ϕj) ⊆ Vαj for every j ∈ Z+.

Thus f =

∞

• j=1

fj has been written as a sum of functions that are C p

• n Ω and spt(fj) ⊆ Vαj for every j ∈ Z+.

So f can be written as a sum of functions fj that are as smooth as f . This allows us to pass from local results to global ones; for example, if we know that a certain property holds on small open sets in Ω, then we can show that a similar property holds on all of Ω by using a partition of unity subordinate to a covering of Ω which consists of small open sets. This technique will be used when working on integration on Rn. Using partitions of unity, the integral can be extended from Jordan regions to open bounded sets even though such sets are not always Jordan regions. This will be the multidimensional version of the improper integrals. It will be used also to prove the change of variables for multiple integrals when the Jacobian of the change of variables transformation is nonzero except on a set of volume zero.

Metric spaces

Most of the properties we have seen for the Euclidean space Rn depend on the metric space structure of Rn, that is, the distance function between any two points. To see the above topological ideas in a metric space, see Chapter 10 Metric Spaces in your textbook. A set X together with a function d : X × X → R is said to be a metric space (that is, the pair (X, d) is said to be a metric space) if the function d is a metric, that is, it satisfies the following three properties: For all x, y, z ∈ X,

• 1. Positive definite: d(x, y) ≥ 0 with d(x, y) = 0 if and only if

x = y.

• 2. Symmetry: d(x, y) = d(y, x).
• 3. Triangle inequality: d(x, y) ≤ d(x, z) + d(z, y).

Using the metric, open balls and then open sets are defined. Defining the collection of open sets means giving the topology.

Rn is a separable complete metric space

The reason why Lindel¨

• f Theorem holds is that Rn has a countable

dense subset (which means in the terminology of metric spaces that Rn with its usual metric is a separable metric space). That means Rn has a subset A such that A is countable and A = Rn. If you look at the proof of Lindel¨

• f theorem, you will see that we

have used the set A = Qn of all points whose coordinates are rational numbers. The set A = Qn is a countable dense subset of

• Rn. Moreover, we have considered the collection U consisting of all
• pen balls whose centers are in A = Qn and whose radii are

rational numbers. This collection U of open balls is a countable collection and suffices to determine the topology of Rn (it is called a basis of the topology of Rn) in the sense that a subset V ⊆ Rn is

• pen if and only if for each x ∈ V there exists an open ball U in

the countable collection U such that x ∈ U ⊆ V . Another important property of Rn is completeness: It is a complete metric space under its usual metric, that is, every Cauchy sequence in Rn is convergent. That property comes from the completeness

• f R: Every Cauchy sequence of real numbers is convergent.

Topological spaces

A set X together with a collection T of subsets of X is said to be a topological space (that is, the pair (X, T ) is said to be a topological space) if the collection T satisfies the following properties:

• 1. ∅ and the whole space X are in T .
• 2. If {Vα}α∈A is an (indexed) collection of elements in T , then

the union

• α∈A

Vα is also in T .

• 3. If r ∈ Z+ and {Vk}r

k=1 is a finite collection of elements in T ,

then the intersection

r

• k=1

Vk is also in T . The elements of T are called open sets in the topological space X. So the only property required in this more general definition is that you just give a collection of sets in a set X, call them open, such that ∅ and X are open, and arbitrary union and finite intersection

• f open sets are open. Guess the definition of the continuity of a

function from a topological space to another topological space.

The Angel of Geometry and the Devil of Algebra

The definition of a topological space is much more general than a metric space. But under some suitable separation axioms and some countability axioms, your topological space will be metrizable, that is, its topology will be the one induced by a metric (see Urysohn metrization theorem in a topology book). You shall investigate these topological properties in general in your course MAT3049 Introduction to Topology next year. The topological terminology that you have seen is just the subject

• f point-set topology. Do not be mislead that topology is just
• these. Better remember homotopic paths, being simply connected
• r Stokes’ theorem or M¨
• bius strip to have some rough idea of the

GEOMETRY-TOPOLOGY field. It is better if you study Differentiable Manifolds to have some idea about the geometry and topology; differential forms will lead you to de Rham cohomology. The Angel of Geometry and the Devil of Algebra fight for the soul of any mathematical being. Attributed to Hermann Weyl

References I

Wade, William R. Introduction to Analysis. 4th edition, Pearson, 2010. Fitzpatrick, P. M. Advanced Calculus. Second Revised edition. American Mathematical Society, 2009. Spivak, Michael. Calculus on manifolds. A modern approach to classical theorems of advanced calculus. W. A. Benjamin, Inc., New York-Amsterdam, 1965. Apostol, Tom M. Mathematical analysis: a modern approach to advanced calculus. Addison-Wesley Publishing Company, Inc., Reading, Mass., 1957. Rudin, Walter. Principles of mathematical analysis. Third

• edition. International Series in Pure and Applied Mathematics.

McGraw-Hill Book Co., New York-Auckland-D¨ usseldorf, 1976.

References II

Apostol, Tom M. Calculus. Vol. I: One-variable calculus, with an introduction to linear algebra. Vol. II: Multi-variable calculus and linear algebra, with applications to differential equations and probability. Second edition. Blaisdell Publishing

• Co. Ginn and Co., Waltham, Mass.-Toronto, Ont.-London,

1967, 1969. Buck, R. Creighton. Advanced calculus. With the collaboration

• f Ellen F. Buck. Third edition. International Series in Pure

and Applied Mathematics. McGraw-Hill Book Co., New York-Auckland-Bogot´ a, 1978. Viro, O. Ya., Ivanov, O. A., Netsvetaev, N. Yu. and Kharlamov, V. M. Elementary Topology: Problem Textbook. American Mathematical Society, 2008. Munkres, James R. Topology. Second edition. Prentice Hall, 2003.

References III

Bredon, Glen E. Topology and Geometry. Springer, 1st ed. 1993, Corr. 3rd printing 1997.