The Urysohn metrization theorem gives conditions which guarantee that a topological space is metrizable. A topological space 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 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 be the sets we want to separate. We can construct a countable open cover of , , whose closures intersect not by taking a open neighborhoods of each element of . 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 to get a similar cover .

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 s and s such that they do not mutually intersect in their closures. A way to do that would be that for any and , we have the part of in subtracted away from it if and also the other way round. This would give us and . ▢

**Urysohn’s lemma** *Let and be disjoint closed sets in a normal space . Then, there is a continuous function such that and .*

*Proof:* Observe that if for all dyadic fractions (those with least common denominator a power of ) , we assign open subsets of such that

- contains and is disjoint from for all
- implies that

and set if for any and otherwise, we are mostly done. Obviously, and . To show that it is continuous, it suffices to show that the preimages of and are open for any . For , the preimage is the union of over , as for any element to go to , by being an infimum, there must be a such that contains it. Now, suppose and take . Then, is an open neighborhood of that maps to a subset of . We see that , with if otherwise, and thereby for and . Moreover, with , we have excluded anything that does not map above .

Now we proceed with the aforementioned assignment of subsets. In the process, we construct another assignment . Initialize and . Let and be disjoint open sets containing and respectively (this is where we need our normality hypothesis). Notice how in normality, we have disjoint closed sets and with open sets and disjoint which contain them respectively, one can complement to derive a closed set larger than , which we call and run the same normal separation process on and . With this, we can construct and the relations

,

.

Inductively, we can show that we can continue this process on and for each provided and on all dyadics with denominator to fill in the ones with denominator . One can draw a picture to help visualize this process and to see that this satisfies the required aforementioned conditions for . ▢

Now we will find a metric for 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 -ball contains some open set of the product space where a cofinite number of the indices project to . The value of for as well as its powers is unbounded, so obviously we need to enforce that the distance exceed not some finite value, say . We also need that for any , 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 th index to , and instead of summing (what would be a series), we take a , which allows for all where , the th index is as desired. We let our metric be

.

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

**Proposition** *The metric induces the product topology on .*

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

,

where is the largest such that . 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 which project to not the entire space but instead to those with length we can assume to be at most . That must have a maximum, which we call . For this we can simply take the minimum over of the length of the interval for divided by as our . ▢

Now we need to construct a homeomorphism from our second-countable, regular (and thereby normal) space to some subspace of . A homeomorphism is injective as part of definition. How to satisfy that? Provide a countable collection of continuous functions to 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 at one and 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 where , we do this on and .

**Proposition** *Our above construction is homeomorphic to .*

*Proof:* Call our function . Each of its component functions is continuous so the entire Cartesian product is also continuous. It remains to show the other way, that in the domain open implies the image of is open. For that it is enough to take for any and find some open neighborhood of it contained in . contains some basis element of the space and thus, there is a component (call it ) that sends to all to and not to . This essentially partitions by vs not , with the latter portion lying inside , which means that is strictly inside . The projections in product space are continuous so that set must be open. This suffices to show that 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.