# 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 $U_i$s and $V_i$s 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 $i$th 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 $n$th 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.