Lecture 5: closed sets, and an introduction to continuous functions Saul Glasman September 16, 2016 Clarification on URL. To warm up today, let s talk about one more example of a topology. Definition 1. Let X be a set. The cofinite topology T fc on X is the following class of subsets: U T fc if and only if U = or X \ U is finite. In order to prove that T fc is a topology, it ll be convenient to introduce closed sets in a topological space. The definition is simple: Definition 2. Let X be a topological space. A set Z X is called closed if its complement Z \ X is open. We ll talk a lot more about closed sets later, but for now you should think of closed sets as sets which are in sharp focus : they have no fuzzy edges. Let s have some examples of closed sets. Example 3. Since X \ = X and X \ X =, and X are always closed. Example 4. In R, the closed interval [a, b] = {r R a r b} is a closed set. To show this, we need to show that the complement R \ [a, b] = (, a) (b, ) = {r R r < a or r > b} is open. But I can write this as a union of open intervals: (, a) = n N(a n, a) and similarly for (b, ). In particular, we can take a = b. Then the closed interval [a, a] is the single point {a}, and so {a} is a closed subset of R. This is very typical behavior: for many nice topological spaces, it s the case that single points are closed subsets. 1
Let s now recall an extra bit of set theory. Lemma 5. Taking complements turns intersections into unions and unions into intersections: Let X be a set and let (U i ) i I be a collection of subsets of X. Then X \ ( U i ) = \ U i ) i I i I(X and X \ ( i I U i ) = i I(X \ U i ). These two statements are known as demorgan s laws. If they re not immediately obvious to you, I recommend you try to prove them at home as an exercise. The upshot is that we can define a topology just as well with closed sets as with open sets: Lemma 6. Let X be a set and let T be a collection of subsets of X. Let T c = {X \ U U T } be the set of complements of elements of T. Then T is a topology if and only if T c satisfies the following conditions: 1. T c. 2. X T c. 3. An arbitrary intersection of elements of T c is in T c. 4. A finite union of elements of T c is in T c. Of course, if you know what the closed sets are, you also know what the open sets are: X \ (X \ U) = U, so (T c ) c = T. The point I want to make right now is that in some cases, such as the cofinite topology, it s easier to check the topology axioms with closed sets than with open sets. Indeed, a subset Z X is closed in the cofinite topology if and only if it s finite or equal to X. But 1. the empty set is finite; 2. X is equal to X; 3. an arbitrary intersection of finite sets is finite; 4. a finite union of finite sets is finite. 2
Although the indiscrete topology is pretty useless, the cofinite topology actually shows up in the wild (for example, in algebraic geometry). It s the coarsest topology which satisfies the reasonable condition that one-point sets are closed. Now let s learn how to relate topological spaces to one another: the study of continuous functions between topological spaces. Show of hands on who s encountered the ɛ δ definition of continuity before. Judging by prerequisites, it should be everyone. To give a brief reminder: a function f : R R is continuous if, informally, when x changes a small amount, f(x) only changes a small amount. This rules out functions with jumps, or crazy functions like { 1 x Q χ Q (x) = 0 x / Q. The actual definition is as follows: Definition 7. f is continuous at x R if for every ɛ > 0, there is some δ > 0 such that f((x δ, x + δ)) (f(x) ɛ, f(x) + ɛ). In other words, if x x < δ, then f(x) f(x ) < ɛ. f is continuous if it s continuous at x for every x R. Our goal for the moment is to rephrase this in a way that makes sense for any function between two topological spaces. Let s start by noting that we can be more flexible with the nature of our intervals: Lemma 8. f is continuous at x if and only if for every open interval set I containing f(x), there s an open interval J containing x such that f(j) I. Proof. Suppose f is continuous. Let I be an open interval with f(x) I. Then there s some ɛ > 0 such that (f(x) ɛ, f(x) + ɛ) I. Then for the corresponding δ whose existence is guaranteed by continuity, we can take J = (x δ, x + δ). Then f(j) I. Conversely, suppose f satisfies our condition. Take ɛ > 0, and let I = (f(x) ɛ, f(x)+ɛ). Then there s some open interval J with x J and f(j) I. There is some δ such that (x δ, x + δ) J. So f is continuous. With this lemma, we ve not only given ourselves more wiggle room and made the definition (I think) much prettier, we ve almost banished the real numbers from the definition statement - it s now a statement solely about open sets and open intervals. Let s make this definition even snappier with the aid of a little piece of language. Definition 9. Let X and Y be sets and f : X Y be a function. If U is a subset of Y, then the preimage of U under f is f 1 (U) = {x X f(x) U}. 3
Don t confuse this notation with the function f 1, the inverse of f when f is bijective! Preimages make perfect sense even if f is not bijective. If f is bijective, then the preimage of U is also the image of U under f 1, so there s no notational ambiguity. Here are some facts about preimages that I ll leave as an exercise: Lemma 10. Let f : X Y be a function. Then If (U i ) i I is a family of subsets of Y, then ( ) f 1 U i = f 1 (U i ). i I i I Similarly, f 1 ( i I U i ) = i I f 1 (U i ). If U, V Y, then f 1 (U \ V ) = f 1 (U) \ f 1 (V ). With this, we can reformulate the definition of continuity in a way that will generalize to topological spaces: Lemma 11. f : R R is continuous if and only if the preimage under f of any open set is open. Proof. To rephrase our previous lemma: f is continuous if and only if for every open set I and every x with f(x) I, there s an open interval J with x J and f(j) I. Rephrasing some more, this is the same as saying: for every x f 1 (I), there s an open interval J with x J and J f 1 (I). But this is the same as saying that f 1 (I) is open. Definition 12. Let X and Y be topological spaces. A function f : X Y is called continuous if the preimage under f of any open subset of Y is an open subset of X. A continuous function is often called a continuous map, or just a map. Remark 13. Since f 1 (Y \ U) = X \ f 1 (U), f is continuous if and only if the preimages under f of closed subsets are closed. It s time for some trivial examples. Example 14. The identity function is always continuous, of course, since id 1 (U) = U. 4
Example 15. Suppose that f : X Y and g : Y Z are continuous functions. Then g f is continuous: indeed, for U Z, (g f) 1 (U) = f 1 (g 1 (U)). But g 1 (U) is open since g is continuous, and so (g f) 1 (U) is open, since f is continuous. Example 16. Let X and X be two topological spaces with the same underlying set. Then id : X X is continuous if and only if the topology on X is at least as fine as the topology on X. Example 17. Suppose X has the discrete topology. Then for any space Y, any function f : X Y is continuous. This makes intuitive sense if you regard the discrete topology as making X into a bunch of loose points. Example 18. Suppose Y has the indiscrete topology. Then for any space X, any function f : X Y is continuous. Example 19. This one isn t quite as immediate as the others, but it s still very easy, and I ll leave the proof as an exercise: if Y has the cofinite topology, then f : X Y is continuous if and only if f 1 {y} is closed for each y Y. 5