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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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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Apr 10, 2012 at 23:38
  • $\begingroup$ @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. $\endgroup$
    – jlk
    Commented Apr 10, 2012 at 23:53
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    $\begingroup$ 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 $\endgroup$
    – David White
    Commented Apr 11, 2012 at 0:47
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    $\begingroup$ 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. $\endgroup$
    – Theo Buehler
    Commented Apr 11, 2012 at 1:26
  • $\begingroup$ @TheoBuehler I'm not sure about "write down", because off-hand it feels like countable choice is used pretty significantly in these existence theorems. $\endgroup$
    – Todd Trimble
    Commented Aug 4, 2018 at 15:58

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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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    $\begingroup$ 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. $\endgroup$
    – Theo Buehler
    Commented Apr 11, 2012 at 1:14
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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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Urysohn's lemma implies that the space of compactly supported continuous functions on a locally compact group is nontrivial, and moreover, rich enough (with supports contained in any neighbourhood of identity, and with functions equal to 1 on any given compact). This allows for the construction of the Haar measure and all harmonic analysis built thereupon.

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What is a good application of Urysohn's Theorem?

This old question was just bumped to the top and caught my attention. Since all the answers given are related to Analysis, let me mention how I apply Urysohn's Theorem when teaching Model Theory.

I cite Urysohn's Theorem to establish that the the space of complete first-order theories in a countable signature $\sigma$ is a Polish space, hence the Cantor-Bendixson Theorem may be applied to it. This shows that if $T$ is a $\sigma$-theory, then the closed subspace $V(T)$ of complete extensions of $T$ uniquely decomposes as $S\cup P$ where $S$ is the countable, scattered subspace of complete theories with ordinal-valued Cantor Bendixson rank and $P$ is the perfect subspace of complete theories of Cantor-Bendixson rank $\infty$. If $P\neq \emptyset$, then $P$ is homemorphic to the Cantor space. These facts may be applied when discussing how difficult it is to axiomatize (relative to $T$) a complete extension of $T$.

Urysohn's Theorem is applied after one has proved the Completeness Theorem and Compactness Theorem, so one already knows that the space of complete $\sigma$-theories is a Stone space. One also knows that the space of complete $\sigma$-theories is second countable when $\sigma$ is countable. One cites Urysohn to prove that a second countable Stone space is a compact metrizable space, hence is a Polish space. It is possible to avoid citing Urysohn's Theorem by (using up a bit more class time and) proving directly that a second countable Stone space is metrizable.

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  • $\begingroup$ Re: the final sentence, isn't it easier than either approach to just give a specific complete metric on the space of complete theories? E.g. fix a listing of the sentences and set $d(T,S)$ to be $2^{-n}$ where $n$ is the index of the first sentence on which $T$ and $S$ differ. $\endgroup$
    – Noah Schweber
    Commented Jun 3 at 4:01
  • $\begingroup$ @NoahSchweber: Dear Noah- I don't disagree with you, but for a simple presentation we have to make a choice about the topology to be used on the space of complete theories, and how to introduce it. I start with the lattice of all $\sigma$-theories, show it to be a spatial frame, and use standard machinery from there. If you want to exhibit an explicit metric on this space, then you have to show that the metric topology is the same as the one associated to the frame. $\endgroup$
    – Keith Kearnes
    Commented Jun 3 at 4:46
  • $\begingroup$ @NoahSchweber: also, the metrizability of the space of theories is only needed to cite the Cantor-Bendixson Theorem in the case of countable language. One can avoid all of this if one does not care that the perfect kernel is either empty or homeomorphic to the Cantor space. If you are satisfied by saying that the perfect kernel is simply "perfect", then the metrizability no longer matters. $\endgroup$
    – Keith Kearnes
    Commented Jun 3 at 4:48

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