EDAA40 EDAA40 Discrete Structures in Computer Science Discrete Structures in Computer Science 4: To infinity and beyond 4: To infinity and beyond Jörn W. Janneck, Dept. of Computer Science, Lund University
one, two, many source: xkcd Science, no. 306, 15 Oct 2004, pp. 496-499 Today's class: learning new numbers, and how to “count” with them. 2
why infinity matters to computing Everything is finite. So are computers. Then why do we care about infinity in maths for computing? Infinity can be used as a model , an abstraction, of two kinds of phenomena: - an awful lot, i.e. very many - the absence of a finite bound Those two things are related, but are not quite the same. 3
Cantor-Schröder-Bernstein theorem When working with infinite stuff, this theorem makes our lives a lot easier. For any two sets A and B, if there are two injections and then there exists a bijection Corollary: If and then . (Proof is a little tricky, we will omit it here. See course page for refs.) 4
Cantor-Schröder-Bernstein theorem Why do we need to prove CSB? Didn't we know this already? What property does this establish for on cardinal numbers? Make sure you clearly distinguish between what is defined, and what needs to be proven. “It’s not what you know, but what you can prove.” Det. Alonzo Harris, LAPD Note: The theorem tells us that there is a bijection. It does not tell us, what it looks like! In other words, it is non-constructive . 5
infinite sets There are various ways to define infinite sets. This is one by Dedekind, 1888: Richard Dedekind 1831-1916 A is infinite if it is equinumerous to a proper subset of itself. That is, there is some S such that Show that the natural numbers are an infinite set. 1. Find a proper subset. 2. Construct a bijection between it and the natural numbers. 6
denumerable (countable) sets A is denumerable ( countable ) if it is equinumerous to the natural numbers, i.e. Denumerable sets are very important to math and CS. They are the smallest infinite sets. (We won't prove that.) The cardinality of the natural numbers (and thus all denumerable sets) has a name: is therefore the smallest transfinite cardinal number . 7
: the integers is the set of integers. -3 -2 -1 0 1 2 3 And this is the bijection: 8.1 Show that z is bijective. 8
products What is the cardinality of ? 9.1 Construct a bijection Make sure you show it is, 9.2 Construct a bijection in fact, bijective in each case! 9
: the rational numbers Some properties: 1. is dense : between any two distinct there is 2. Any non-empty open interval is equinumerous to -3 -2 -1 0 1 2 3 For example: 10.1 Show that is bijective. 10.2 Show that 2. above is true for all open intervals. This is simpler than you might think. 10
: the rational numbers Let's start with the simple stuff, i.e. Then we define another set, this one of pairs of integers and natural numbers: Now, there is clearly (*) a bijection between and : So those two sets are equinumerous. Now it's also the case that Put all this together: (*) In everyday parlance, words like “clearly”, “of course”, “obviously” etc. are used to tell you that what follows requires no further inspection or reflection. In math, think of them as technical expressions that mean just the opposite: pay EXTRA attention! 11
finite sequences/strings Let A be a finite set of n symbols . The set of all finite sequences (strings) of these symbols is The empty sequence is . What is if A is... 12
infinite sequences Let A be a finite set of n symbols . An infinite sequence in A is a function The set of all infinite sequences in A: As always: 13
Cantor's diagonal construction 1. Let's start by assuming that , i.e. there must be a bijection . Recall that a bijection is also surjective, i.e. Georg Cantor 2. Assuming an f, we can construct the 1845-1918 diagonal sequence D: ... 0 1 2 3 4 5 6 7 8 9 3. Invert D: 0 1 0 0 1 1 1 0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 2 0 0 1 1 0 1 1 0 1 0 4. Note that 3 1 1 1 0 1 0 0 1 0 0 ... 4 0 1 0 1 0 1 0 1 0 1 5 1 1 0 0 1 1 0 0 1 1 5. This violates the assumption 6 0 1 0 1 0 1 0 0 1 0 that f is a bijection. 7 0 0 0 0 1 1 1 0 0 0 Conclusion: 1 8 1 1 0 0 0 1 1 1 0 There is no bijection 9 0 1 1 0 0 0 1 1 1 1 . . . ... ... 14
more than To summarize: 1. We have 2. … but we cannot construct a bijection 3. Conclusion: We discovered a new cardinal number: Proposition: It is the case that for all finite 15.1 Show 1. by constructing an injection 15.2 Show that for any finite , n > 1 (the “proposition”). Tip: Use the Cantor-Schröder-Bernstein theorem. 15
programs and functions boolean f (BigInteger n) { … } How many programs implementing f? How many functions ? Conclusion? 16
: the real numbers -3 -2 -1 0 1 2 3 So for the purposes of determining the cardinality of the real numbers, we can focus on non-terminating decimal sequences: Non-terminating means there is no n such that all digits after are 0. Otherwise, this would be the set of all sequences of ten symbols , with cardinality Even so, the cardinality of the real numbers still comes out to (proof deferred) 17
power sets Note that for , it is the case that , i.e. the set of natural numbers is strictly smaller than its powerset. This holds more generally: For any set A, same thing For any cardinal number C, How do we show this? 1. Obviously, 2. We need to show they aren't equinumerous. Let's suppose they are, i.e. there must be a bijection 3. Construct the “diagonal set” . Note that 4. f is surjective, so there must be an x such that 5. If then either or Conclusion: There is no such x, so f can never be surjective thus What does that mean for transfinite cardinal numbers? 18
more transfinite cardinals Paul Cohen 1934-2007 So far, we have encountered two transfinite cardinals: and . s As we have seen, there are infinitely many transfinite cardinals. i Starting from , they are called in order d Such that between any two there is no other cardinal number. Where does fit in? All we know is that , so it's at least . e So, is ? This is the continuum hypothesis (CH) . b CH was shown to be independent of ZFC (Cohen, 1963). Since ZFC doesn't tell us how big those alephs are, we get beths: a Such that and . At least we know that r Note: We assume ZFC for this discussion, i.e. Zermelo-Fraenkel set theory with the axiom of choice. Do not worry about it. 19
proof for (sketched) Let's show that . s First, let's say every real number between 0 and 1 corresponds to a sequence in this set: i (*) d but Note that Our hypothesis is that which means e Constructing the bijection is very tedious. Instead, using the CSB theorem, we can simply construct two injections: b a 20.1 Finish the proof that by constructing those two injections. r Tip: Start with g. That one is really easy. * Recall that , so it's the set of all decimal digits. 20
Recommend
More recommend
