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Definitions:

Let $X$ denote a topological space throughout.

If all singleton subsets of $X$ are closed, then we call $X$ Fréchet. If, given any closed subset $C \subset X$ and any point $x \in X - C$, there exist disjoint neighbourhoods of $x$ and $C$, then we call $X$ quasiregular. If $X$ is both Fréchet and quasiregular, then we call $X$ regular. (To--hopefully--avoid confusion, I forgo the use of the $T_n$ notation for separation properties entirely, and use the conventions of Clark's notes on general topology in this PDF.

A collection $\mathcal{A}$ of subsets of $X$ is said to be locally finite if, for every point $x \in X$, there exists some neighbourhood of $x$ which intersects only finitely-many elements of $\mathcal{A}$. If we can write a collection $\mathcal{B}$ of subsets of $X$ as a countable union $\bigcup_{n \in \mathbb{N}} \mathcal{A}_n$, where each $\mathcal{A}_n$ is locally finite, then the collection $\mathcal{B}$ is said to be $\sigma$-locally finite.


My question:

I am confused about the proof of one direction of the Nagata-Smirnov metrisation theorem given as Lemma 4.20 in Kelley, which for completeness I attach to this post as an image. In the above terminology, the statement of this lemma is

A regular space whose topology is generated by a $\sigma$-locally finite basis is metrisable.

As I understand it, his approach is to make use of the $\sigma$-locally finite basis to define a countable collection $\{ d_{(n,m)} \}_{(n,m) \in \mathbb{N}^2}$ of continuous functions $X \to \mathbb{R}$, and then show that this collection distinguishes points from closed subsets of $X$. This allows us to deduce that the evaluation map $x \mapsto \left( d_{(n,m)}(x) \right)_{(n,m) \in \mathbb{N}^2}$ embeds $X$ as a subspace of the metrisable space $\prod_{(n,m) \in \mathbb{N}^2} \mathbb{R}$, which in turn tells us that $X$ is itself metrisable. (Morally, this seems basically identical to the standard proof of the same direction of Urysohn's metrisation theorem, modulo the technical details of defining the countable collection of functions $X \to \mathbb{R}$ in the first place.)

Now, there exist metrisable spaces which fail to be separable (or, equivalently, fail to be second countable); for instance, the discrete topology on any uncountable underlying set. What I don't understand is why Lemma 4.20 does not imply that all metrisable spaces are separable when we combine it with the reverse implication.

Explicitly, if we assume that the topology of any metrisable space is generated by some $\sigma$-locally finite basis, then my understanding of Lemma 4.20 tells us that we can embed such spaces in the countable product $\prod_{n \in \mathbb{N}} \mathbb{R}$. As a product of countably-many second countable spaces, $\prod_{n \in \mathbb{N}} \mathbb{R}$ is second countable; since second countability is inherited by subspaces, this seems to imply the (false) result that all metrisable spaces are second countable.

Where is my understanding going wrong?


Lemma 4.20 of Kelley's *General Topology*. Where Kelley uses the term "regular", I use "quasiregular"; where Kelley uses <span class=$T_1$, I use "Fréchet".">

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  • $\begingroup$ A space whose singletons are closed is usually called a $T_1$ space and a Frechet space is usually defined as something else. E.g. see Engelking, General Topology. $\endgroup$ – DanielWainfleet Apr 30 '19 at 17:53
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It's comparable to the write-up in my note (note the similarity to 4.14 in Kelley):

Lemma 3: Let $X$ be regular and $T_0$ and let $d_i, i \in \mathbb{N}$ be a countable family of continuous pseudometrics on $X$ such that they are all bounded by $1$ and such that for every closed $A$ and every $x \notin A$, there is some $i$ such that $d_i(x,A)>0$, where $d_i(x,A)=\inf\{d_i(x,a): a \in A\}$ as usual. Then $X$ is metrisable and $$d(x,y)=\sum_i \frac{1}{2^i}d_i(x,y) $$ is a compatible metric.

So it's not an embedding argument but a direct construction of the metric from the countably many continuous pseudometrics. The countability of the family comes from the "$\sigma$" in $\sigma$-locally finite base, and the well-definedness of each pseudometric from the local finiteness, if you analyse the proof.

You can prove the lemma by the embedding argument, as Kelley does: let $X_n$ be $X$ in the topology induced by the pseudometric $d_n$, the continuity of the pseudometric implies that the identity mapping $i_n$ from $X$ to $X_n$ is continuous. A standard fact is that the weighted sum pseudometric induces the product topology on $\prod_n X_n$. The property of the family of pseudometrics then implies the family of $i_n$ separates points and closed sets and the standard embedding theorem then gives the stated result that the $d$ metric I defined works. So we don’t embed into a countable product of copies of reals but in a countable product of pseudometric spaces of the same size as $X$. Not separable necessarily.

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The book wrote:

There is a countable family $D$ such that for each closed $A$ and $x\notin A$, there is a pseudometric $d_n\in D$ from $X\times X$ to $\Bbb R$ such that $d_n(x,A)>0$.

Thus the embedding should be

$$f: X\rightarrow \Bbb R^J \text{ where } J=\Bbb N\times\mathscr A \text{ and } \mathscr A=\{A\subseteq X\mid A \text{ closed }\}$$

$$x\mapsto (d_n(x,A))_{(n,A)\in J }$$

Clearly, $J$ need not be countable, so $\Bbb R^J$ need not be second countable.

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  • $\begingroup$ That's not what they wrote, though. Read again.... $\endgroup$ – Henno Brandsma Apr 30 '19 at 13:20
  • $\begingroup$ @HennoBrandsma The first sentence of the proof of lemma 20? $\endgroup$ – YuiTo Cheng Apr 30 '19 at 13:24
  • $\begingroup$ No, the family does not depend on $A$. $\endgroup$ – Henno Brandsma Apr 30 '19 at 13:25
  • $\begingroup$ @HennoBrandsma I don't think so. Quote: "There is a countable family $D$...such that for each closed subset $A$ of $X$...there is a member $d$ of $D$..." $\endgroup$ – YuiTo Cheng Apr 30 '19 at 13:27
  • $\begingroup$ Indeed, logically: $\exists D : \forall A: \exists d \in D$ so one family works for all $A$.. You're claiming $\forall A: \exists D$ etc, which is not what is intended. $\endgroup$ – Henno Brandsma Apr 30 '19 at 13:29

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