Urysohn metrization theorem

The Urysohn metrization theorem gives conditions which guarantee that a topological space is metrizable. A topological space (X, \mathcal{T}) is metrizable is there is a metric that induces a topology that is equivalent to the topological space itself. These conditions are that the space is regular and second-countable. Regular means that any combination of closed subset and point not in it is separable, and second-countable means there is a countable basis.

Metrization is established by embedding the topological space into a metrizable one (every subspace of a metrizable space is metrizable). Here, we construct a metrization of [0,1]^{\mathbb{N}} and use that for the embedding. We first prove that regular and second-countable implies normal, which is a hypothesis of Urysohn’s lemma. We then use Urysohn’s lemma to construct the embedding.

Lemma Every regular, second-countable space is normal.

Proof: Let B_1, B_2 be the sets we want to separate. We can construct a countable open cover of B_1, \{U_i\}, whose closures intersect not B_2 by taking a open neighborhoods of each element of B_1. With second-countability, the union of those can be represented as a union of a countable number of open sets, which yields our desired cover. Do the same for B_2 to get a similar cover \{V_i\}.

Now we wish to minus out from our covers in such a way that their closures are disjoint. We need to modify each of the U_is and V_is such that they do not mutually intersect in their closures. A way to do that would be that for any U_i and V_j, we have the part of \bar{U_i} in V_j subtracted away from it if j \geq i and also the other way round. This would give us U_i' = U_i \setminus \sum_{j=1}^i \bar{V_j} and V_i' = V_i \setminus \sum_{j=1}^i \bar{V_j}.     ▢

Urysohn’s lemma Let A and B be disjoint closed sets in a normal space X. Then, there is a continuous function f : X \to [0,1] such that f(A) = \{0\} and f(B) = \{1\}.

Proof: Observe that if for all dyadic fractions (those with least common denominator a power of 2) r \in (0,1), we assign open subsets of X U(r) such that

  1. U(r) contains A and is disjoint from B for all r
  2. r < s implies that \overline{U(r)} \subset U(s)

and set f(x) = 1 if x \notin U(r) for any r and f(x) = \inf \{r : x \in U(r)\} otherwise, we are mostly done. Obviously, f(A) = \{0\} and f(B) = \{1\}. To show that it is continuous, it suffices to show that the preimages of [0, a) and (a, 1] are open for any x. For [0, a), the preimage is the union of U(r) over r < a, as for any element to go to a' < a, by being an infimum, there must be a s \in (a', a) such that U(s) contains it. Now, suppose f(x) \in (a, 1] and take s \in (a, f(x)). Then, X \setminus \bar{U(s)} is an open neighborhood of x that maps to a subset of (a, 1]. We see that x \in X \setminus \overline{U(s)}, with if otherwise, s < f(x) and thereby f(x) \leq s' < f(x) for s' > s and U(s') \supset \overline{U(s)}. Moreover, with s > a, we have excluded anything that does not map above a.

Now we proceed with the aforementioned assignment of subsets. In the process, we construct another assignment V. Initialize U(1) = X \setminus B and V(0) = X \setminus A. Let U(1/2) and V(1/2) be disjoint open sets containing A and B respectively (this is where we need our normality hypothesis). Notice how in normality, we have disjoint closed sets B_1 and B_2 with open sets U_1 and U_2 disjoint which contain them respectively, one can complement B_1 to derive a closed set larger than U_2, which we call U_2' and run the same normal separation process on A_1 and U_2'. With this, we can construct U(1/4), U(3/4), V(1/4), V(3/4) and the relations

X \setminus V(0) \subset U(1/4) \subset X \setminus V(1/4) \subset U(1/2),

X \setminus U(1) \subset V(3/4) \subset X \setminus U(3/4) \subset V(1/2).

Inductively, we can show that we can continue this process on X \setminus V(a/2^n) and X \setminus U((a+1)/2^n) for each a = 0,1,\ldots,2^n-1 provided U and V on all dyadics with denominator 2^n to fill in the ones with denominator 2^{n+1}. One can draw a picture to help visualize this process and to see that this satisfies the required aforementioned conditions for U.     ▢

Now we will find a metric for \mathbb{R}^{\mathbb{N}} the product space. Remember that the base for product space is such that all projections are open and a cofinite of them are the full space itself (due to closure under only finite intersection). Thus our metric must be such that every \epsilon-ball contains some open set of the product space where a cofinite number of the indices project to \mathbb{R}. The value of x - y for x,y \in \mathbb{R} as well as its powers is unbounded, so obviously we need to enforce that the distance exceed not some finite value, say 1. We also need that for any \epsilon > 0, the distance contributed by all of the indices but a finite number exceeds it not. For this, we can tighten the upper bound on the ith index to 1/i, and instead of summing (what would be a series), we take a \sup, which allows for all n > N where 1/N < \epsilon, the nth index is \mathbb{R} as desired. We let our metric be

D(\mathbf{x}, \mathbf{y}) = \sup\{\frac{\min(|x_i-y_i|, 1)}{i} : i \in \mathbb{N}\}.

That this satisfies the conditions of metric is very mechanical to verify.

Proposition The metric D induces the product topology on \mathbb{R}^{\mathbb{N}}.

Proof: An \epsilon-ball about some point must be of the form

(x_1 - \epsilon/2, x_1 + \epsilon/2) \times (x_2 - 2\epsilon/2, x_2 + 2\epsilon/2) \times \cdots \times (x_n - n\epsilon/2, x_n + n\epsilon/2) \times \mathbb{R} \times \cdots \times \mathbb{R} \times \cdots,

where n is the largest such that n\epsilon < 1. Clearly, we can fit into that an open set of the product space.

Conversely, take any open set and assume WLOG that it is connected. Then, there must be only a finite set of natural number indices I which project to not the entire space but instead to those with length we can assume to be at most 1. That must have a maximum, which we call n. For this we can simply take the minimum over i \leq n of the length of the interval for i divided by i as our \epsilon.     ▢

Now we need to construct a homeomorphism from our second-countable, regular (and thereby normal) space to some subspace of \mathbb{R}^\mathbb{N}. A homeomorphism is injective as part of definition. How to satisfy that? Provide a countable collection of continuous functions to \mathbb{R} such that at least one of them differs whenever two points differ. Here normal comes in handy. Take any two distinct points. Take two non-intersecting closed sets around them and invoke Urysohn’s lemma to construct a continuous function. That would have to be 0 at one and 1 at the other. Since our space is second-countable, we can do that for each pair of points with only a countable number. For every pair in the basis B_n, B_m where \bar{B_n} \subset B_m, we do this on \bar{B_n} and X \setminus B_m.

Proposition Our above construction is homeomorphic to [0,1]^{\mathbb{R}}.

Proof: Call our function f. Each of its component functions is continuous so the entire Cartesian product is also continuous. It remains to show the other way, that U in the domain open implies the image of U is open. For that it is enough to take z_0 = f(x_0) for any x_0 \in U and find some open neighborhood of it contained in f(U). U contains some basis element of the space and thus, there is a component (call it f_n) that sends X \setminus U to all to 0 and x_0 not to 0. This essentially partitions X by 0 vs not 0, with the latter portion lying inside U, which means that \pi_n^{-1}((0, \infty)) \cap f(X) is strictly inside f(U). The projections in product space are continuous so that set must be open. This suffices to show that f(U) is open.     ▢

With this, we’ve shown our arbitrary regular, second-countable space to be homeomorphic to a space we directly metrized, which means of course that any regular, second-countable space is metrizable, the very statement of the Urysohn metrication theorem.

Another characterization of compactness

The canonical definition of compactness of a topological space X is every open cover has finite sub-cover. We can via contraposition translate this to every family of open sets with no finite subfamily that covers X is not a cover. Not a cover via de Morgan’s laws can be characterized equivalently as has complements (which are all closed sets) which have finite intersection. The product is:

A topological space is compact iff for every family of closed sets with the finite intersection property, the intersection of that family is non-empty.