Lecture 3: the classification of equivalence relations and the definition of a topological space Saul Glasman September 12, 2016 If there s a bijection f : X Y, we ll often say that X and Y are in bijection, and we ll act kind of as if X and Y are the same, since the elements of Y are in one-to-one correspondence with the elements of X; they re just the elements of X under different names. Now on the first day I promised a result on equivalence relations; today I ll be able to give the statement and the proof. After that, we ll get started on topological spaces. Let X be a set. We can form the set of all equivalence relations on X - equivalence relations are things, we can put them in a set. Call this set Eq(X). Definition 1. A partition of X is a collection (X i ) i I of subsets of X such that if i j, then X i X j =, and the union X i = X. i I In other words, each x X is in exactly one of the X i. Don t be scared of the notation (X i ) i I ; the point is that we have a big list of subsets of X, and we have to give them names, so we index them with tags from some convenient index set I. It doesn t matter exactly what I is; it just has to be the right size of set to index our collection. Now we can form the set of all partitions of X - call it Part(X). Then we have the following theorem: Theorem 2. The sets Eq(X) and Part(X) are in bijection. Remark 3. I m showing you this theorem partly because it s interesting, and partly to demonstrate that set theory can get meta : questions about the nature of sets can themselves be treated using set theory. This theorem and its proof will serve as a warm-up for the kind of level of abstraction we ll be encountering in this course. 1
Proof. In view of the theorem I proved last time, if I claim two sets are in bijection, there are two ways I can prove that: I can give a function one way and show that it s injective and surjective, or I can give functions both ways and show they re inverse to one another. Here we ll do the latter. First I ll prove some auxiliary results. If R is an equivalence relation on X and x X, then the equivalence class of x, EC R (x) X is defined as the set of x X such that xrx. Lemma 4. If xrx, then EC R (x ) = EC R (x). Proof. Suppose z EC R (x). Then xrx and x Rz. By transitivity, which was one of our axioms for equivalence relations, xrz. We deduce that EC R (x ) EC R (x). But by symmetry, which was another of our axioms, we have x Rx. So the same reasoning shows that EC R (x) EC R (x ). Each subset is included in the other, so the two must be equal. by Now let s define φ : Eq(X) Part(X) φ(r) = {EC R (x) x X}. Notice we ve overcounted a lot here: each equivalence class is counted separately for each element of that equivalence class. But that doesn t matter. We first have to prove that φ(r) is really a partition; otherwise the codomain of φ isn t what I claimed it is. That is, we have to show that each element x X is in exactly one equivalence class. But if an equivalence class C contains x, then by the lemma, C must be the equivalence class of x, so x is in at most one equivalence class. Since xrx, x is in its own equivalence class. We deduce that x is in exactly one equivalence class, and φ(r) is really a partition. Now define ψ : Part(X) Eq(X) by setting, for a partition ψ(p) = R P, where P = {(X i ) i I } xr P x if x X i, x X i for some i. Now we have to show that R P is an equivalence relation. But I actually gave this proof last week when I proved that landmasses give an equivalence relation! So I ll omit this in lecture, and for those of you who weren t here on the first day I ll leave it as an exercise, but for completeness, I ll give it here. 2
1. Each x is in some partition X i. So xr P x. 2. If x X i, y X i, and y X j, z X j, then i = j, because P is a partition. But then z X i, so xr P z. 3. If x X i, y X i, then y X i, x X i. So now we have our well-defined functions and φ : Eq(X) Part(X) ψ : Part(X) Eq(X). We ll now show that φ and ψ are mutually inverse. This is actually really clear, once you unpack the notation, but I ll go through it for completeness. First, let s show that ψ φ = id. (I ll stop writing the subscript on id when the context is clear.) This is the same as saying that by taking the equivalence relation associated to the equivalences classes, you get the same equivalence relation - that is, xrx if and only if they re in the same equivalence class for R. That, we already know. Now let s show that φ ψ = id. This is the same as saying that the equivalence classes for the equivalence relation associated to a partition are exactly the sets of that partition - that is, x and x are in the same equivalence class for ψ(p) if and only if there is some X i for which both x X i, x X i. This, too, we know; it s just the definition of ψ. So we re done! Finally. I apologize for the tedium and the length of the proof, but hopefully each step was clear individually. If you didn t follow the overall structure of the proof, I highly recommend you look over it in the notes; there will be plenty of proofs in the course which are at least this complicated. Now let s really begin the course! This subject truly starts with the definition of a topological space, and that s what we re about to cover. First a little bit of background. Topology grew out of the study of continuous functions. You ve probably encountered the definition of a continuous function from R to R using δs and ɛs (if not, don t worry, you will soon.) This definition is fine, but it does rely on the real numbers having a well-defined notion of distance - that is, the absolute difference between two real numbers. As mathematics grew in abstraction, the need for an idea of continuity that would generalize to contexts where notions of distance were unavailable or irrelevant became clear. The crucial insight was that the definition of continuity could be reformulated in terms of open sets - we ll see how this works within the next few classes. This led to an abstract, set-theoretic definition of topological space making reference only to a class of open sets. Before giving this definition, let me informally set up some intuition about what an open set should be in the context of R. I m not going to give you the precise definition of an open set just yet. Loosely speaking, an open set is a set 3
without any sharp edges - a set you can t cut yourself on. The archetypal example of an open set is the open interval (a, b) = {x R a < x < b}. Another example of an open set is the entire set R. Sets which are not open include the half-open interval (a, b] = {x R a < x b}, which has one sharp edge, and the closed interval [a, b] = {x R a x b}, which has two sharp edges. With this in mind, let s get to the definition of a topology: Definition 5. Let X be a set. A topology on X is a collection T of subsets of X, known as open sets satisfying the following conditions: 1. T. 2. X T. 3. Any union of elements of T is in T. 4. A finite intersection of elements of T is in T. Why do we allow arbitrary unions but only finite intersections? Consider the following example: If n is a positive integer, let I n R be the open interval (0, 1 + 1 n ). Now let s take the intersection of all of these open sets: I = I n. n=1 What are the elements of I? Well, they re the real numbers which are greater than 0 but which are less than 1 + 1/n for all n. So certainly, Also, 1 I. But if r > 1 - say (0, 1) I. r = 1.00000000000001 - then there s some n such that r > 1 + 1/n, right? So I = (0, 1]. Oops - that s not open any more. If we d only taken a finite intersection, though, it would have been fine. 4
So that s the definition of a topology. The pair (X, T ) is called a topological space. Let s give a few easy examples. Example 6. Let X be a set. Then the discrete topology on X is the set of all subsets of X: T = P(X) (P(X) is the notation for the set of all subsets of X; it s called the power set, in case you didn t know.) It s clear that the discrete topology satisfies the axioms for a topology. Example 7. Let X be a set. Then the indiscrete topology on X is the topology for which only and X are open: T = {, X}. Again, it s clear that the indiscrete topology satisfies the axioms for a topology: the only nontrivial collection of elements of T is {, X}, and X = X, X =. Let s give the definition of an open subset of R. Definition 8. Let U R. We say U is open if for every r U, r is contained in an open interval which is contained in U; that is, for every r in U, there are positive numbers ɛ 0, ɛ 1 > 0 such that (r ɛ 0, r + ɛ 1 ) U. Proposition 9. With this definition, the open subsets of R form a topology on R. We ll deduce this result as a special case of a more general theorem next time. 5