Characterizing Endpoints of Generalized Inverse Limits Lori Alvin - - PowerPoint PPT Presentation

Characterizing Endpoints of Generalized Inverse Limits

Lori Alvin

Bradley University

Nipissing Topology Workshop 2018

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 1 / 44

Motivation

There has long been a push to understand the topological structure of inverse limit spaces generated by unimodal maps. Although such maps are simple to define, their inverse limits can exhibit quite complicated structures, and it is often difficult to distinguish the inverse limits associated with distinct maps.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 2 / 44

Motivation

There has long been a push to understand the topological structure of inverse limit spaces generated by unimodal maps. Although such maps are simple to define, their inverse limits can exhibit quite complicated structures, and it is often difficult to distinguish the inverse limits associated with distinct maps. One of the things we have focused on in the study of these classical inverse limits is the set of endpoints, as they are a topological invariant. Thus, it is logical to investigate the properties of endpoints and use them to explore and distinguish inverse limits.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 2 / 44

Motivation

There has long been a push to understand the topological structure of inverse limit spaces generated by unimodal maps. Although such maps are simple to define, their inverse limits can exhibit quite complicated structures, and it is often difficult to distinguish the inverse limits associated with distinct maps. One of the things we have focused on in the study of these classical inverse limits is the set of endpoints, as they are a topological invariant. Thus, it is logical to investigate the properties of endpoints and use them to explore and distinguish inverse limits. As we have transitioned to the study of generalized inverse limits, it remains helpful to study the collection of endpoints; however they can arise differently from set-valued functions than from continuous maps.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 2 / 44

Motivation

In this talk, we will demonstrate how the collection of endpoints arises from set-valued inverse limits by looking at various projection mappings.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 3 / 44

Motivation

In this talk, we will demonstrate how the collection of endpoints arises from set-valued inverse limits by looking at various projection mappings. We note that although the standard definitions of endpoints of continua are equivalent when working with chainable continua, because our inverse limits can contain triods, we must be careful which definition we work

• with. In fact, characterizations of endpoints under one definition may not

hold in our setting under a different definition.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 3 / 44

Motivation

In this talk, we will demonstrate how the collection of endpoints arises from set-valued inverse limits by looking at various projection mappings. We note that although the standard definitions of endpoints of continua are equivalent when working with chainable continua, because our inverse limits can contain triods, we must be careful which definition we work

• with. In fact, characterizations of endpoints under one definition may not

hold in our setting under a different definition. After presenting several examples of endpoints for generalized inverse limits satisfying a given property, we discuss our conjecture for the characterization of endpoints and the progress/stumbling blocks in proving the conjecture.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 3 / 44

Generalized Inverse Limits

Let X be a continuum; 2X is the space of all non-empty, compact subsets

• f X.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 4 / 44

Generalized Inverse Limits

Let X be a continuum; 2X is the space of all non-empty, compact subsets

• f X.

Given a function F : X → 2X we define its graph to be the set Γ(F) = {(x, y) : y ∈ F(x)}.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 4 / 44

Generalized Inverse Limits

Let X be a continuum; 2X is the space of all non-empty, compact subsets

• f X.

Given a function F : X → 2X we define its graph to be the set Γ(F) = {(x, y) : y ∈ F(x)}. lim ← − F =

• x ∈

∞

• i=0

X : xi−1 ∈ F(xi) for all i ∈ N

• Lori Alvin (Bradley U.)

Endpoints lalvin@bradley.edu 4 / 44

Generalized Inverse Limits

Let X be a continuum; 2X is the space of all non-empty, compact subsets

• f X.

Given a function F : X → 2X we define its graph to be the set Γ(F) = {(x, y) : y ∈ F(x)}. lim ← − F =

• x ∈

∞

• i=0

X : xi−1 ∈ F(xi) for all i ∈ N

• Γ1 = X and for all n ≥ 2 we define the projection Γn by:

Γn =

• x ∈

n−1

• i=0

X : xi−1 ∈ F(xi) for all 1 ≤ i < n

• Lori Alvin (Bradley U.)

Endpoints lalvin@bradley.edu 4 / 44

Inverse and Forward Limits

One of the keys to understanding lim ← − F is to look at the graph of F −1. If Γ(F) = {(x, y) : y ∈ F(x)}, then Γ(F −1) = {(y, x) : y ∈ F(x)}. We are interested in set-valued functions F : [0, 1] → 2[0,1] such that Γ(F −1) = ∪α∈AΓ(fα), where each fα : [0, 1] → [0, 1] is a continuous function. lim ← − F = lim − → F −1 = lim − →(∪α∈Afα) So really, these techniques can be used to study the forward dynamics of set-valued functions.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 5 / 44

Definitions of Endpoints

Bing’s Definition: The point p is an endpoint of the continuum X if for any two subcontinua H, K ⊆ X both containing p, either H ⊆ K or K ⊆ H.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 6 / 44

Definitions of Endpoints

Bing’s Definition: The point p is an endpoint of the continuum X if for any two subcontinua H, K ⊆ X both containing p, either H ⊆ K or K ⊆ H. Lelek’s Definition The point p is an endpoint of the continuum X if and

• nly if p is an endpoint of every arc in X that contains p.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 6 / 44

Definitions of Endpoints

Bing’s Definition: The point p is an endpoint of the continuum X if for any two subcontinua H, K ⊆ X both containing p, either H ⊆ K or K ⊆ H. Lelek’s Definition The point p is an endpoint of the continuum X if and

• nly if p is an endpoint of every arc in X that contains p.

Miller’s Definition The point p is an endpoint in the continuum X if every irreducible continuum in X containing p is irreducible between p and some other point.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 6 / 44

Motivating Question

Theorem (J. Kelly (2016))

Let F : X → 2X. Suppose there exists a collection {fα}α∈A of continuous functions such that Γ(F −1) = ∪α∈AΓ(fα). Then for every p ∈ lim ← − F the following are equivalent using Bing’s definition of an endpoint.

1 p is an endpoint of lim

← − F.

2 (p0, p1, . . . , pn−1) is an endpoint of Γn for infinitely many n ∈ N. 3 (p0, p1, . . . , pn−1) is an endpoint of Γn for all n ∈ N.

Note that it always follows (even with Lelek’s and Miller’s definitions) that if p is an endpoint of lim ← − F, then (p0, . . . , pn−1) is an endpoint of Γn for all n ∈ N.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 7 / 44

Motivating Question Does Kelly’s result hold if we assume Lelek’s (Miller’s) definition of an endpoint? If not, what is the proper characterization for Lelek’s (Miller’s) definition of an endpoint in this setting?

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 8 / 44

Our Counterexample: F

Let F : [0, 1] → 2[0,1] be the set-valued function obtained by attaching the line segment connecting the points ( 1

2, 1+ √ 5 4

) and (1, 1) to the tent map Ts with slope s = 1+

√ 5 2

.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 9 / 44

Our Counterexample: F

Let F : [0, 1] → 2[0,1] be the set-valued function obtained by attaching the line segment connecting the points ( 1

2, 1+ √ 5 4

) and (1, 1) to the tent map Ts with slope s = 1+

√ 5 2

.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 9 / 44

Our Counterexample: F −1 = f ∪ g

We can decompose F −1 into the union of two functions F −1 = f ∪ g: f g f (x) =

• 2x

1+ √ 5

x ≤ 1+

√ 5 4

,

2x+1− √ 5 3− √ 5

x ≥ 1+

√ 5 4

. g(x) = −2x+1+

√ 5 1+ √ 5

x ≤ 1+

√ 5 4

,

2x+1− √ 5 3− √ 5

x ≥ 1+

√ 5 4

.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 10 / 44

Our Counterexample: Γ2

(1, 1) (0, 1) (0, 0) ( 1+

√ 5 4

, 1

2)

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 11 / 44

Our Counterexample: Γ3

(1, 1, 1) (0, 1, 1) (0, 0, 0) ( 1+

√ 5 4

, 1

2, √ 5−1 4

) ( 1+

√ 5 4

, 1

2, 5− √ 5 4

) (0, 0, 1) (

√ 5−1 4

, 1+

√ 5 4

, 1

2)

( 3

√ 5−3 4

, 1+

√ 5 4

, 1

2)

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 12 / 44

Our Counterexample: Γ4

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 13 / 44

Our Counterexample: Γ2

(1, 1) (0, 1) (0, 0) ( 1+

√ 5 4

, 1

2)

Notice that Γ2 is the union of the arcs (x0, f (x0)) := (f ) and (x0, g(x0)) := (g) where x0 ∈ [0, 1]. These two arcs coincide on (x0, f (x0)) = (x0, g(x0)) when x0 ∈ [ 1+

√ 5 4

, 1]. We call the point ( 1+

√ 5 4

, 1

2) a

branch point of Γ2.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 14 / 44

Our Counterexample: Γ3

(1, 1, 1) (0, 1, 1) (0, 0, 0) ( 1+

√ 5 4

, 1

2, √ 5−1 4

) ( 1+

√ 5 4

, 1

2, 5− √ 5 4

) (0, 0, 1) (

√ 5−1 4

, 1+

√ 5 4

, 1

2)

( 3

√ 5−3 4

, 1+

√ 5 4

, 1

2)

Notice that Γ3 is the union of four arcs: (f , f ), (f , g), (g, g), (g, f ). There are four branch points in Γ3. We can use the branch points in Γn to define the branch points of lim ← − F. Issue: There is a sequence of branch points in lim ← − F that converge to (1, 1, 1, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 15 / 44

Our Counterexample: lim − → f and lim − → g

f g Note that lim ← − F contains both lim − → f and lim − → g. lim − → f is an arc with endpoints (0, 0, 0, . . .) and (1, 1, 1, . . .). lim − → g is an arc with endpoint (0, 1, 1, . . .) and (1, 1, 1, . . .). We show that lim − → f ∩lim − → g = {(1, 1, 1, . . .)} and thus lim − → f ∪lim − → g is an arc.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 16 / 44

Our Counterexample: (1, 1, 1, . . .) NOT endpoint

f g Let x ∈ lim − → f and y ∈ lim − → g such that x, y = (1, 1, 1, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 17 / 44

Our Counterexample: (1, 1, 1, . . .) NOT endpoint

f g Let x ∈ lim − → f and y ∈ lim − → g such that x, y = (1, 1, 1, . . .). Then gn(y0) → m (attracting fixed point for g) and f n(x0) → 0 (attracting fixed point for f ).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 17 / 44

Our Counterexample: (1, 1, 1, . . .) NOT endpoint

f g Let x ∈ lim − → f and y ∈ lim − → g such that x, y = (1, 1, 1, . . .). Then gn(y0) → m (attracting fixed point for g) and f n(x0) → 0 (attracting fixed point for f ). Thus the coordinates of x, y must eventually disagree. Hence x = y.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 17 / 44

Our Counterexample: (1, 1, 1, . . .) NOT endpoint

f g Let x ∈ lim − → f and y ∈ lim − → g such that x, y = (1, 1, 1, . . .). Then gn(y0) → m (attracting fixed point for g) and f n(x0) → 0 (attracting fixed point for f ). Thus the coordinates of x, y must eventually disagree. Hence x = y. So lim − → f ∩ lim − → g = {(1, 1, 1, . . .)}. Hence lim − → f ∪ lim − → g is an arc with endpoints (0, 0, 0, . . .) and (0, 1, 1, . . .) that contains (1, 1, 1, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 17 / 44

Our Counterexample: (1, 1, 1, . . .) NOT endpoint

f g Let x ∈ lim − → f and y ∈ lim − → g such that x, y = (1, 1, 1, . . .). Then gn(y0) → m (attracting fixed point for g) and f n(x0) → 0 (attracting fixed point for f ). Thus the coordinates of x, y must eventually disagree. Hence x = y. So lim − → f ∩ lim − → g = {(1, 1, 1, . . .)}. Hence lim − → f ∪ lim − → g is an arc with endpoints (0, 0, 0, . . .) and (0, 1, 1, . . .) that contains (1, 1, 1, . . .). (1, 1, 1, . . .) is NOT an endpoint of lim ← − F under Lelek’s definition.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 17 / 44

Our Counterexample: Any Endpoints?

(1, 1, 1) (0, 1, 1) (0, 0, 0) ( 1+

√ 5 4

, 1

2, √ 5−1 4

) ( 1+

√ 5 4

, 1

2, 5− √ 5 4

) (0, 0, 1) (

√ 5−1 4

, 1+

√ 5 4

, 1

2)

( 3

√ 5−3 4

, 1+

√ 5 4

, 1

2)

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 18 / 44

Our Counterexample: Any Endpoints?

(1, 1, 1) (0, 1, 1) (0, 0, 0) ( 1+

√ 5 4

, 1

2, √ 5−1 4

) ( 1+

√ 5 4

, 1

2, 5− √ 5 4

) (0, 0, 1) (

√ 5−1 4

, 1+

√ 5 4

, 1

2)

( 3

√ 5−3 4

, 1+

√ 5 4

, 1

2)

Similar arguments will show that (0, 1, 1, . . .), (0, 0, 1, 1, . . .), etc. will not be endpoints of lim ← − F, as there will always be a sequence of branch points that converges to those points.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 18 / 44

Our Counterexample: Any Endpoints?

(1, 1, 1) (0, 1, 1) (0, 0, 0) ( 1+

√ 5 4

, 1

2, √ 5−1 4

) ( 1+

√ 5 4

, 1

2, 5− √ 5 4

) (0, 0, 1) (

√ 5−1 4

, 1+

√ 5 4

, 1

2)

( 3

√ 5−3 4

, 1+

√ 5 4

, 1

2)

Similar arguments will show that (0, 1, 1, . . .), (0, 0, 1, 1, . . .), etc. will not be endpoints of lim ← − F, as there will always be a sequence of branch points that converges to those points. The only endpoint of lim ← − F is (0, 0, 0, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 18 / 44

Comparison to lim ← − Ts

Recall that the tent map Ts is such that its turning point is periodic with period 3. Thus lim ← − Ts has the following set of endpoints:

• (0, 0, 0, . . .),
• 1

2, 1 + √ 5 4 , √ 5 − 1 4 , . . .

• ,
• 1 +

√ 5 4 , √ 5 − 1 4 , 1 2, . . .

• ,

√ 5 − 1 4 , 1 2, 1 + √ 5 4 , . . . . Somehow attaching the arc to the symmetric tent map removes all of these endpoints except (0, 0, 0, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 19 / 44

Endpoints for a Generalized Family Fs, s > 1

Let Fs be the set-valued function obtained from Ts by attaching a straight line from the critical point to (1, 1). For all 1 < s < 2, the only endpoint of lim ← − Fs is (0, 0, 0, . . .). We also do not know what lim ← − Fs actually looks like, because it contains lim ← − Ts.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 20 / 44

Endpoints for a Generalized Family Fs, s = 1

f g f (x) = x g(x) =

• 1 − x

x ≤ 1

2,

x x ≥ 1

2

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 21 / 44

Endpoints for a Generalized Family Fs, s = 1

(0, 0) (0, 1) ( 1

2, 1 2)

(1, 1) Γ2 (0, 0, 0) (0, 1, 1) ( 1

2, 1 2, 1 2)

(1, 1, 1) Γ3 (0, 0, 1) Γ4 lim ← − F1

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 22 / 44

Endpoints for a Generalized Family Fs, s = 1

The collection of endpoints are all points of the form: (1, 1, 1, . . .), (0, 0, 0, . . .) or (0, . . . , 0, 1, 1, . . .). The inverse limit is a countable fan.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 23 / 44

Endpoints for a Generalized Family Fs, s < 1

The collection of endpoints are all points of the form: (1, 1, 1, . . .), (0, 0, 0, . . .) or (0, . . . , 0, 1, 1, . . .). The inverse limit is a comb; all such inverse limits are homeomorphic.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 24 / 44

Endpoints for a Generalized Family Hs, s > 1

Let Hs be the set-valued function obtained from Ts by attaching a straight line from the critical point to ( 1

2, 1).

For all 1 < s < 2, the collection of endpoints is {p : pi = 1 for some i ∈ N} ∪ {(0, 0, 0, . . .)}.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 25 / 44

Endpoints for a Generalized Family Hs, s > 1

We can decompose H−1

s

into the union of two functions H−1

s

= f ∪ g: f g f (x) =

• x

s

x ≤ s

2, 1 2

x ≥ s

2.

g(x) =

• 1 − x

s

x ≤ s

2, 1 2

x ≥ s

2.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 26 / 44

Endpoints for a Generalized Family Hs, s > 1: Γ2

(1, 1

2)

(0, 1) (0, 0) ( s

2, 1 2)

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 27 / 44

Endpoints for a Generalized Family Hs, s > 1: Γ3

(1, 1

2, 1 2s )

(0, 1, 1

2)

(0, 0, 0) ( s

2, 1 2, 1 2s )

(1, 1

2, 2s−1 2s )

( s

2, 1 2, 2s−1 2s )

(0, 0, 1) ( 2s−s2

2

, s

2, 1 2)

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 28 / 44

Endpoints for a Generalized Family Hs, s > 1

Note that regardless of the 1 < s < 2, every point in the inverse limit space which contains a 1 will be an endpoint. (0, 0, 0, . . .) is also an endpoint.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 29 / 44

Endpoints for a Generalized Family Hs, s > 1

Note that regardless of the 1 < s < 2, every point in the inverse limit space which contains a 1 will be an endpoint. (0, 0, 0, . . .) is also an endpoint. This is because every pair of arcs that contains such a point p will necessarily overlap on a nondegenerate arc, forcing the point p to be an endpoint of each of those arcs.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 29 / 44

Endpoints for a Generalized Family Hs, s > 1

Note that regardless of the 1 < s < 2, every point in the inverse limit space which contains a 1 will be an endpoint. (0, 0, 0, . . .) is also an endpoint. This is because every pair of arcs that contains such a point p will necessarily overlap on a nondegenerate arc, forcing the point p to be an endpoint of each of those arcs. The branch points in this case DO NOT approach the endpoints.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 29 / 44

Endpoints for a Generalized Family Hs, s = 1

The collection of endpoints are all points of the form: (0, 0, 0, . . .), (0, 0, . . . , 0, 1, 1

2, 1 2, . . .) or (1, 1 2, 1 2, . . .).

The inverse limit is a countable fan, which is homeomorphic to lim ← − F1.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 30 / 44

Endpoints for a Generalized Family Hs, s < 1

The collection of endpoints are all points of the form: (0, 0, 0, . . .), (0, 0, . . . , 0, 1, 1

2, 1 2, . . .) or (1, 1 2, 1 2, . . .).

The inverse limit is a comb; all such inverse limits are homeomorphic, and they are also homeomorphic to lim ← − Fs.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 31 / 44

Endpoints for a Generalized Family Gs

Let Gs be the set-valued function obtained from Ts by attaching a straight line from the critical point to (0, 1). This case is very complicated! Of the three families we studied, this is the

• nly one where the collection of endpoints completely depends on the slope

s (that is, it doesn’t just break into the s = 1, s < 1 and s > 1 cases).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 32 / 44

Decompose Gs, s > 1

We can decompose G −1

s

into the union of two functions G −1

s

= f ∪ g: f g f (x) =

• x

s

x ≤ s

2, 1−x 2−s

x ≥ s

2.

g(x) =

• 1 − x

s

x ≤ s

2, 1−x 2−s

x ≥ s

2.

f has an attracting fixed point at 0. g has an attracting fixed point at

s s+1.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 33 / 44

Observations about f and g

If x > s

2, then f (x) = g(x).

g(x) > s

2 if and only if x < 2s−s2 2

. f (x) < s

2 for all x.

Therefore, for any point (x0, x1, x2, . . .) ∈ lim ← − Gs, if xi ∈ [ s

2, 1], then

xi = g(xi−1) and xi+1 = f (xi) = g(xi). Further, it is not possible for xixi+1 = 11. The set of possible endpoints of lim ← − Gs is the set P = {p ∈ {0, 1}N : pipi+1 = 11}.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 34 / 44

Any point with tail 010101 . . . is NOT an endpoint

Note that lim ← − Gs contains lim − → g ∪ lim − →(g, f , g, f , . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 35 / 44

Any point with tail 010101 . . . is NOT an endpoint

Note that lim ← − Gs contains lim − → g ∪ lim − →(g, f , g, f , . . .). lim − → g is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 1, 0, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 35 / 44

Any point with tail 010101 . . . is NOT an endpoint

Note that lim ← − Gs contains lim − → g ∪ lim − →(g, f , g, f , . . .). lim − → g is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 1, 0, . . .). lim − →(g, f , g, f , . . .) is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 0, 1, 0, 1, 0, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 35 / 44

Any point with tail 010101 . . . is NOT an endpoint

Note that lim ← − Gs contains lim − → g ∪ lim − →(g, f , g, f , . . .). lim − → g is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 1, 0, . . .). lim − →(g, f , g, f , . . .) is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 0, 1, 0, 1, 0, . . .). If 0 < x0 < 2s−s2

2

, then g2(x0) = f (g(x0)) =

x0 2s−s2 > x0.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 35 / 44

Any point with tail 010101 . . . is NOT an endpoint

Note that lim ← − Gs contains lim − → g ∪ lim − →(g, f , g, f , . . .). lim − → g is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 1, 0, . . .). lim − →(g, f , g, f , . . .) is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 0, 1, 0, 1, 0, . . .). If 0 < x0 < 2s−s2

2

, then g2(x0) = f (g(x0)) =

x0 2s−s2 > x0.

As

s s+1 is an attracting fixed point of g, there will exist an even n such

that gn(x0) > 2s−s2

2

.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 35 / 44

Any point with tail 010101 . . . is NOT an endpoint

Note that lim ← − Gs contains lim − → g ∪ lim − →(g, f , g, f , . . .). lim − → g is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 1, 0, . . .). lim − →(g, f , g, f , . . .) is an arc with endpoints (0, 1, 0, 1, . . .) and (1, 0, 0, 1, 0, 1, 0, . . .). If 0 < x0 < 2s−s2

2

, then g2(x0) = f (g(x0)) =

x0 2s−s2 > x0.

As

s s+1 is an attracting fixed point of g, there will exist an even n such

that gn(x0) > 2s−s2

2

. Thus gn+1(x0) = f (gn(x0)).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 35 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .). WLOG, suppose x0 = y0 < 2s−s2

2

.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .). WLOG, suppose x0 = y0 < 2s−s2

2

. Then there will exist some n such that xn = yn; thus x = y.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .). WLOG, suppose x0 = y0 < 2s−s2

2

. Then there will exist some n such that xn = yn; thus x = y. The only point in lim − → g ∩ lim − →(g, f , g, f , . . .) is (0, 1, 0, 1, . . .).

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .). WLOG, suppose x0 = y0 < 2s−s2

2

. Then there will exist some n such that xn = yn; thus x = y. The only point in lim − → g ∩ lim − →(g, f , g, f , . . .) is (0, 1, 0, 1, . . .). Hence (0, 1, 0, 1, . . .) is NOT an endpoint of lim ← − Gs.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .). WLOG, suppose x0 = y0 < 2s−s2

2

. Then there will exist some n such that xn = yn; thus x = y. The only point in lim − → g ∩ lim − →(g, f , g, f , . . .) is (0, 1, 0, 1, . . .). Hence (0, 1, 0, 1, . . .) is NOT an endpoint of lim ← − Gs. In fact, any point with tail 0101 · · · will NOT be an endpoint of lim ← − Gs.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

Any point with tail 010101 . . . is NOT an endpoint

Let x ∈ lim − → g with x = (0, 1, 0, 1, . . .) and y ∈ lim − →(g, f , g, f , . . .) with y = (0, 1, 0, 1, . . .). WLOG, suppose x0 = y0 < 2s−s2

2

. Then there will exist some n such that xn = yn; thus x = y. The only point in lim − → g ∩ lim − →(g, f , g, f , . . .) is (0, 1, 0, 1, . . .). Hence (0, 1, 0, 1, . . .) is NOT an endpoint of lim ← − Gs. In fact, any point with tail 0101 · · · will NOT be an endpoint of lim ← − Gs. Similar arguments show that for each 1 < s < 2 there exists an Ns such that if the number of consecutive 0’s in the tail of p ∈ P is less than Ns, then p is not an endpoint of lim ← − Gs.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 36 / 44

General Facts about Endpoints of lim ← − Gs

If p ∈ P has only finitely many 1’s, then p will be an endpoint.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 37 / 44

General Facts about Endpoints of lim ← − Gs

If p ∈ P has only finitely many 1’s, then p will be an endpoint.

1 There is an N ∈ N such that pi = 0 for all i ≥ N, so pi+1 = f (pi). 2 The only branch points occur when xn = s

2, but f (xi) < 1 2 for all

xi ∈ [0, 1].

3 There is an arc containing p and exactly one branch point. 4 Thus every arc in lim

← − Gs with p has p as an endpoint.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 37 / 44

General Facts about Endpoints of lim ← − Gs

If p ∈ P has only finitely many 1’s, then p will be an endpoint.

1 There is an N ∈ N such that pi = 0 for all i ≥ N, so pi+1 = f (pi). 2 The only branch points occur when xn = s

2, but f (xi) < 1 2 for all

xi ∈ [0, 1].

3 There is an arc containing p and exactly one branch point. 4 Thus every arc in lim

← − Gs with p has p as an endpoint. For every s > 1, there is an Ns ∈ N such that if the number of consecutive 0’s in the tail of p ∈ P is always greater than Ns, then p will be an endpoint of lim ← − Gs.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 37 / 44

General Facts about Endpoints of lim ← − Gs

If p ∈ P has only finitely many 1’s, then p will be an endpoint.

1 There is an N ∈ N such that pi = 0 for all i ≥ N, so pi+1 = f (pi). 2 The only branch points occur when xn = s

2, but f (xi) < 1 2 for all

xi ∈ [0, 1].

3 There is an arc containing p and exactly one branch point. 4 Thus every arc in lim

← − Gs with p has p as an endpoint. For every s > 1, there is an Ns ∈ N such that if the number of consecutive 0’s in the tail of p ∈ P is always greater than Ns, then p will be an endpoint of lim ← − Gs. If p ∈ P has fewer than Ns consecutive 0’s infinitely often in it is tail, then p may or may not be an endpoint of lim ← − Gs; it depends on the pattern of 1’s and 0’s immediately following each such occurrence.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 37 / 44

Endpoints for a Generalized Family Gs, s = 1

The collection of endpoints is the Cantor set of points in {0, 1}N such that no two 1’s appear consecutively. The inverse limit is a Cantor fan.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 38 / 44

Endpoints for a Generalized Family Gs, s < 1

The collection of endpoints is the Cantor set of points in {0, 1}N such that no two 1’s appear consecutively. The inverse limit is such that each branch has many branches, which has many branches, etc; further, all such inverse limits are homeomorphic.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 39 / 44

Current Work

We are currently working on getting a better understanding of endpoints for the family of functions F : [0, 1] → 2[0,1] such that Γ(F −1) = ∪α∈Afα.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 40 / 44

Current Work

We are currently working on getting a better understanding of endpoints for the family of functions F : [0, 1] → 2[0,1] such that Γ(F −1) = ∪α∈Afα. Right now, I am looking at extending work containing these symmetric tent families: Allowing the attached arc to connect the critical point to the point (a, 1) (a = 0, 1

2, 1).

Allowing multiple attached arcs to connect to the critical point. Working with other piecewise linear set-valued functions.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 40 / 44

Characterization Conjecture

In all of the examples we worked with, the point p = (p0, p1, p2, . . .) was an endpoint of lim ← − F if either

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 41 / 44

Characterization Conjecture

In all of the examples we worked with, the point p = (p0, p1, p2, . . .) was an endpoint of lim ← − F if either

1 There is NO sequence of branch points converging to p; or Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 41 / 44

Characterization Conjecture

In all of the examples we worked with, the point p = (p0, p1, p2, . . .) was an endpoint of lim ← − F if either

1 There is NO sequence of branch points converging to p; or 2 The sequence of branch points converging to p also corresponds to a

sequence of other endpoints converging to p.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 41 / 44

Characterization Conjecture

Conjecture: Let F : [0, 1] → 2[0,1] be a set-valued function such that there exists a collection of continuous functions {fα : [0, 1] → [0, 1]} such that Γ(F −1) = ∪α∈AΓ(fα). Then for every p ∈ lim ← − F, the following are equivalent using Lelek’s definition of an endpoint.

1 p is an endpoint of lim

← − F.

2 (p0, p1, . . . , pn−1) is an endpoint of Γn for all n ∈ N and either ◮ The distance between (p0, p1, . . . , pn−1) and its closest branch point

(x0, x1, . . . , xn−1) is bounded as n → ∞; or

◮ The distance between (p0, p1, . . . , pn−1) and its closest branch point

(x0, x1, . . . , xn−1) is not bounded, but the other endpoint (q0, q1, . . . , qn−1) with shared branch point (x0, x1, . . . , xn−1) is arbitrarily close to (p0, p1, . . . , pn−1) for large enough n.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 42 / 44

Summary

We have successfully classified the endpoints for several families of set-valued functions obtained from the symmetric tent family.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 43 / 44

Summary

We have successfully classified the endpoints for several families of set-valued functions obtained from the symmetric tent family. We noticed certain behaviors arising in the sequence of Γn’s that allow a point p ∈ lim ← − F to be an endpoint for these families of functions F.

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 43 / 44

Summary

We have successfully classified the endpoints for several families of set-valued functions obtained from the symmetric tent family. We noticed certain behaviors arising in the sequence of Γn’s that allow a point p ∈ lim ← − F to be an endpoint for these families of functions F. Do these patterns generalize to all set-valued functions satisfying Γ(F −1) = ∪α∈AΓ(fα)? (Maybe this will depend if A is finite or infinite)

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 43 / 44

Summary

We have successfully classified the endpoints for several families of set-valued functions obtained from the symmetric tent family. We noticed certain behaviors arising in the sequence of Γn’s that allow a point p ∈ lim ← − F to be an endpoint for these families of functions F. Do these patterns generalize to all set-valued functions satisfying Γ(F −1) = ∪α∈AΓ(fα)? (Maybe this will depend if A is finite or infinite) Is there a way to symbolically characterize the collection of endpoints of lim ← − F? Might such a symbolic characterization rely on {fα}?

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 43 / 44

Thank you!

Lori Alvin (Bradley U.) Endpoints lalvin@bradley.edu 44 / 44