1 Lecture 13: October 8 Urysohn’s metrization theorem. Today, I want to explain some applications of Urysohn’s lemma. The first one has to do with the problem of characterizing metric spaces among all topological spaces. As we know, every metric space is also a topological space: the collection of open balls Br(x) is a basis for the metric topology. A natural question is exactly which topological spaces arise in this way. Definition 13.1. A topological space X is called metrizable if it is homeomorphic to a metric space (with the metric topology). In fact, the answer is known: the Nagata-Smirnov metrization theorem gives a necessary and sufficient condition for metrizability. If you are interested, please see Chapter 6 in Munkres’ book; to leave enough time for other topics, we will not discuss the general metrization theorem in class. We will focus instead on the special case of second countable spaces, which is all that one needs in practice. Recall from last week that every metric space is normal. It turns out that if we restrict our attention to spaces with a countable basis, then normality is equivalent to metrizability; this is the content of the following theorem by Urysohn. In fact, it is enough to assume only regularity: by Theorem 11.1, every regular space with a countable basis is normal. Theorem 13.2 (Urysohn’s metrization theorem). Every regular space with a count- able basis is metrizable. Example 13.3. For example, every compact Hausdor↵space with a countable basis is metrizable: the reason is that every compact Hausdor↵space is normal. There are two ways to show that a given space X is metrizable: one is to construct a metric that defines the topology on X; the other is to find an embedding of X into a metric space, because every subspace of a metric space is again a metric space. To prove the metrization theorem, we first show that the product space [0, 1]! is metrizable (by constructing a metric), and then we show that every regular space with a countable basis can be embedded into [0, 1]! (by using Urysohn’s lemma). Proposition 13.4. The product space [0, 1]! is metrizable. ! Proof. We write the points of [0, 1] in the form x =(x1,x2,...), where 0 xk 1. Recall from our discussion of the product topology that the collection of open sets y [0, 1]! x y <r for k =1,...,n 2 | k − k| k ! is a basis for the topology T ;here x [0, 1] is any point, n 1 is an integer, 2 ≥ and r1,...,rn are positive real numbers. Each of the basic open sets involves only finitely many coordinates; our task is to find a metric in which the open balls have the same property. We define a candidate metric on the product space by the formula 1 1 d(x, y)=sup xk yk = max xk yk . k k | − | k k | − | 1 Because xk yk 1, the numbers k xk yk approach 0 as k gets large; the supremum| is− therefore| achieved for some| k.− It is| an easy exercise to check that d is indeed a metric. To prove that the metric topology Td is the same as the product topology T , we have to compare the basic open sets in both topologies. 2 Let us first show that every open ball Br(x) is open in the product topology. By definition, we have B (x)= y [0, 1]! d(x, y) <r r 2 ! = y [0, 1] xk yk <kr for every k 2 | − | ! 1 = y [0, 1] xk yk <krfor every k r− ; 2 | − | the last equality holds because xk yk <kris automatically satisfied once kr > 1, due to the fact that x ,y [0|, 1].− So| the open ball B (x) is actually a basic open k k 2 r set in the product topology, which means that T T . d ✓ To prove that the two topologies coincide, let U T be an arbitrary open set in the product topology. For any point x U, some2 basic open set 2 y [0, 1]! x y <r for k =1,...,n 2 | k − k| k must be contained in U.Ifwenowdefine r r =min k , 1 k n k then we have kr < rk for every k =1,...,n, and so the open ball B (x)= y [0, 1]! x y <krfor every k r 2 | k − k| is contained in U. This is enough to conclude that U Td, and hence that Td = T ; ! 2 thus the product space [0, 1] is indeed metrizable. ⇤ One consequence is that R! is also metrizable: it is homeomorphic to (0, 1)!, which is a subspace of the metrizable space [0, 1]!. You can see from the proof that the index set really had to be countable in order to define the metric; in fact, one can show that the product of uncountably many copies of R is no longer metrizable. Proof of Urysohn’s metrization theorem. We are now ready to prove Theo- rem 13.2. Let X be a regular space with a countable basis; let me remind you that X is automatically normal (by Theorem 11.1). The idea of the proof is to construct an embedding F : X [0, 1]!,F(x)= f (x),f (x),... ! 1 2 Recall that an embedding is an injective function f : X Y that induces a homeo- morphism between X and the subspace f(X) of Y . Since! a function into a product space is continous if and only if all the coordinate functions f = p F are con- n n ◦ tinuous, we have to look for countably many continuous functions fn : X [0, 1]. There are two other requirements: (1) F should be injective, meaning that! when- ever x = y, there should exist some n with fn(x) = fn(y). (2) F should induce a homeomorphism6 between X and F (X), meaning that6 whenever U X is open, F (U) should be open in the subspace topology on F (X). ✓ Lemma 13.5. There are countably many continuous functions f : X [0, 1] with n ! the following property: for every open set U X and for every point x0 X, there is some index n such that f (x )=1, but f ✓(x)=0for all x U. 2 n 0 n 62 Proof. Since X is normal, we can easily find such a function for every U and every x0: we only have to apply Urysohn’s lemma to the two closed sets x0 and X U. The resulting collection of functions is not going to be countable in{ general,} but\ we can use the existence of a countable basis to make it so. 3 Let B1,B2,... be a countable basis for the topology on X. For every pair of indices m, n such that Bm Bn, the two closed sets Bm and X Bn are disjoint; Urysohn’s lemma gives us a✓ continuous function g : X [0, 1]\ with m,n ! 1ifx Bm, gm,n(x)= 2 0ifx B . 62 n In this way, we obtain countably many continuous functions; I claim that they have the desired property. In fact, suppose that we have an open set U X and a point ✓ x0 U. Because the Bn form a basis,® there is some index n with x Bn U. Now2 X is regular, and so we can find a smaller open set V with 2 ✓ x V V B ; 2 ✓ ✓ n we can also choose another index m such that x B V .ThenB V B , 2 m ✓ m ✓ ✓ n and the function gm,n from above satisfies gm,n(x0) = 1 (because x0 Bm), and g (x) = 0 for x X U (because X U X B ). This shows that the2 countably m,n 2 \ \ ✓ \ n many functions gm,n do what we want. ⇤ Using the functions from the lemma, we can now define a function F : X [0, 1]!,F(x)= f (x),f (x),... ; ! 1 2 because all the individual functions fn are continuous, the function F is also con- tinuous (by Theorem 4.16). Our goal is to show that F is an embedding. Injectivity is straightforward: Suppose we are given two points x, y X with x = y.Thenx belongs to the open set X y , and so Lemma 13.5 guarantees2 that there6 is some \{ } index n for which fn(x) = 1 and fn(y) = 0. But then F (x) = F (y), because their n-th coordinates are di↵erent. 6 This already proves that F is a continuous bijection between X and F (X); the following lemma shows that it is even a homeomorphism. Lemma 13.6. F is a homeomorphism between X and the subspace F (X) of [0, 1]!. Proof. We have to show that for any nonempty open set U X, the image F (U) is open in the subspace topology on F (X). Consider an arbitrary✓ point t F (U); 0 2 note that t0 = F (x0) for a unique x0 U. According to Lemma 13.5, there is some index n with f (x ) = 1 and f (x) =2 0 for every x U. Now consider the set n 0 n 62 W = F (X) t [0, 1]! p (t) > 0 . \ 2 n It is clearly open in F (X), being the intersection of F (X) with a basic open set in [0, 1]!. I claim that t W and W F (U). In fact, we have 0 2 ✓ pn(t0)=pn F (x0) = fn(x0) > 0, which shows that t0 W .