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Urysohn's Metrization Theorem states that every Hausdorff second-countable regular space is metrizable.

What is an example of a Hausdorff second-countable regular space where it is difficult to prove metrizability without using Urysohn's Theorem?

For example, the theorem implies that a (second-countable) manifold is metrizable. However, this result can be proven without using Urysohn's Theorem by showing directly that every such manifold embeds in $\mathbb{R}^{\infty}$ (using partitions of unity).

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I don't know if the metrization theorem itself is very interesting from an "application" perspective. But the Urysohn Lemma used to prove the theorem, that's interesting and has plenty of uses throughout topology. –  Ryan Budney Apr 10 '12 at 23:38
    
@Ryan Budney - I almost thought of asking about Urysohn's Lemma instead of Urysohn's Theorem. I would be interested to hear applications of the Lemma as well. I can ask that as a separate question if you'd prefer. –  jlk Apr 10 '12 at 23:53
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This doesn't answer your question, but it's worth noting that the proof of Urysohn's Theorem easily gives what Munkres calls the Imbedding Theorem (34.2), since you end up imbedding X into some giant $\mathbb{R}^J$. This characterizes completely regular spaces as subspaces of compact Hausdorff spaces. In turn, that theorem is used to prove the Nagata-Smirnov metrization theorem, which actually classifies metric spaces. To me, that's reason enough to develop Urysohn's theorem, but I'll look through my old notes to see if it's ever needed on its own (without Nagata-Smirnov) to get metrizability –  David White Apr 11 '12 at 0:47
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I'm not sure if the question is well-posed: The proof of Urysohn's metrization theorem provides you with a more or less explicit metric coming from an embedding into a product space (the metric looks similar to what I wrote in a comment to an answer) and is related to what you described as a way to circumvent Urysohn's theorem when proving metrizability of manifolds: if you can prove your space to be Hausdorff, regular and second countable then you can write down a metric for its topology. Same goes for Bing-Nagata-Smirnov that was mentioned by David White. –  Theo Buehler Apr 11 '12 at 1:26

2 Answers 2

One good use is to conclude that the unit ball of the dual of a separable Banach space, in the w*-topology, is a compact metric space.

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I very much prefer to write down a metric explicitly (choose a dense sequence $x_n$ of the Banach space and put $d(\varphi,\psi) = \sum_n 2^{-n} \min{\lbrace 1,|\varphi(x_n) - \psi(x_n)|\rbrace}$ for $\phi,\psi$ in the unit ball); then it is straightforward to check that it induces the weak$^\ast$-topology. A diagonal argument shows that the unit ball is compact The advantage of this is that you can teach the Alaoglu theorem to people who aren't acquainted with topology beyond metric spaces. –  Theo Buehler Apr 11 '12 at 1:14

A very remarkable and classical result that uses repeatedly the Urysohn's lemma (not the metrization theorem) is the proof of Riesz representation theorem in its general setting.

The theorem states that if $(X,\tau)$ is a locally compact Hausdorff space, then for every positive linear functional $\Lambda : C_{0}\to \mathbb{R}$ there exists a Borel measure $\mu$ so that for all $f\in C_{0}$:

$$\Lambda (f) = \int_{X} f d \mu$$

Where $C_{0}$ is the collection of all compactly supported real-valued continuous functions on $X$. Conversely, every functional defined as above for a given measure is linear and positive, so this theorem gives one-to-one correspondence with Borel measures and linear positive functionals.

The proof relies explicitly on Urysohn's lemma and partitions of unity that are obtained aswell by using this lemma. Note that a locally compact Hausdorff space is $T_{3}$.

You can find the detailed proof if you're interested from: Rudin W., Real and Complex analysis, 1970, Theorem 2.14

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