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Let us recall that a symmetric on a set $X$ is any function $d:X\times X\to[0,\infty)$ such that for every $x,y\in X$ the following two conditions are satisfied:

$\bullet$ $d(x,y)=0$ if and only if $x=y$;

$\bullet$ $d(x,y)=d(y,x)$.

A topological space $X$ is called symmetrizable if there exists a symmetric $d$ on $X$ such that a subset $U\subseteq X$ is open if and only if for every $x\in X$ there exists $\varepsilon>0$ such that $B_d(x,y)\subseteq U$, where $B_d(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}$.

By the Urysohn Metrization Theorem, every regular second-countable space is metrizable and hence symmetrizable.

Question 1. Is each Hausdorff second-countable space symmetrizable?

A weaker version of Question 1 is also of interest:

Question 2. Is each countable first-countable Hausdorff space symmetrizable?


Added in Edit. I have just realized that Question 2 has a simple affirmative answer (which I present below), so only Question 1 remains open.

Added in Edit 31.05.2022. Ups: Question 1 has a negative answer!

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    $\begingroup$ "A symmetric"? this sounds (to me — not native) as phoney English. $\endgroup$
    – YCor
    May 30 at 17:55
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    $\begingroup$ @YCor Nonetheless this is a standard terminology, see e.g. Section 9 of Gruenhages' survey paper "Generalized metric spaces" in Handbook of Set-Theoretic Topology. $\endgroup$ May 30 at 18:01
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    $\begingroup$ Arhangelskii's 1966 paper Mappings and spaces (this is the Russian Mathematical Surveys translation) might be useful. Gruenhage has a more recent survey, although I suppose you know of it. FYI, I know very little about this topic. I simply looked up the term in this book (p. 237, right column), where I saw the Arhangelskii reference, which I surprisingly have a photocopy of, so I very briefly glanced at it. $\endgroup$ May 30 at 19:40
  • $\begingroup$ @DaveLRenfro You are right. A more general result was proved by Akhangelski in his paper in Russian Math Surveys: according to his Theorem 2.9, every first-countable $T_1$-space with $\sigma$-discrete network is symmetrizable. Thank you for this reference to Arhangelski. I had to look at it before asking this question. $\endgroup$ May 30 at 19:59
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    $\begingroup$ @Ycor Perhaps it's a pun on the word "metric", so one should read it as "a sym-metric"? $\endgroup$ Jun 1 at 8:38

1 Answer 1

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I have just realized that the first question has a simple affirmative answer.

Theorem 1. Every countable first-countable $T_1$ space $X$ is symmetrizable.

Proof. For every $x\in X$, fix a neighborhood base $(U_n(x))_{n\in\omega}$ such that $U_{n+1}(x)\subseteq U_n(x)$ for all $n\in\omega$. Since $X$ is countable, there exists a linear order $\le$ on $X$ such that for every $x\in X$ the initial interval $\{y\in X:y\le x\}$ is finite.

It is easy to see that the topology of $X$ is generated by the symmetric $$d(x,y)=\inf\big\{2^{-n}:\max\{x,y\}\in U_n(\min\{x,y\})\big\}.\quad\square$$


Important Remark. After looking at a reference Dave L. Renfro suggested in a comment as possibly being relevant, I discovered that affirmative answers to both my questions follow from

Theorem 2.9 (Arhangelski, 1966): Every first-countable $T_1$ space with a $\sigma$-discrete (closed) network is symmetrizable.

But reading the proof of this theorem I discovered that it works only for regular spaces. Arhangelski writes that one loses no generality assuming that the $\sigma$-discrete network is closed, i.e., consists of closed sets. But after taking the closure of elements of a network in a $T_1$-space (even in a Hausdorff space), the network property can be destroyed. So, Question 2 seems to stay open and not answered even for Hausdorff (not mention $T_1$) spaces.


Added in Edit 31.05.2022. To my big surprise (and contrary to what was claimed by Arhangelski in his Theorem 2.9), I have just discovered that Question 1 has negative answer!

First let us prove that the symmetrizability of second-countable spaces is equivalent to the perfectness. Recall that a topological space $X$ is perfect if every closed subset of $X$ is of type $G_\delta$.

Theorem 2. A second-countable (Hausdorff) $T_1$-space $X$ is symmetrizable if (and only if) it is perfect.

Proof. If $X$ is symmetrizable and Hausdorff, then the first-countability of $X$ implies that $X$ is semi-metrizable and perfect, see the paragraph before Theorem 9.8 in Gruenhage's "Generalized metric spaces".

Conversely, if a second-countable $T_1$-space $X$ is perfect, then each open subset of $X$ is an $F_\sigma$ in $X$, which implies that $X$ has a countable closed network. Now we can apply Arhangelski's Theorem 2.9 to conclude that $X$ is symmetrizable. $\square$

Let $\mathrm{non}(\mathcal M)$ denote the smallest cardinality of a nonmeager set in the real line.

Example 1. There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable.

Proof. Take any nonmeager linear subspace $L$ of cardinality $\mathrm{non}(\mathcal M)$ in $\mathbb R^\omega$ such that for every $n\in\omega$ the intersection $L_n=L\cap(\{0\}^n\times\mathbb R^{\omega\setminus n})$ is dense in $\{0\}^n\times\mathbb R^{\omega\setminus n}$. Consider the quotient space $X=L_\circ/_\sim$ of $L_\circ=L\setminus\{0\}$ by the equivalence relation $\sim$ defined by $x\sim y$ iff $\mathbb R x=\mathbb Ry$. Since the space $L_\circ$ is Baire and the quotient map $q:L_\circ\to X$ is open, the space $X$ is second-countable and Baire. It is easy to check that the closure of every nonempty set in $X$ contains the set $q[L_n\setminus\{0\}]$ for some $n\in\omega$. This implies that the space $X$ is superconnected in the sense that for every nonempty open sets $U_1,\dots,U_n$ in $X$ the intersection of their closures $\overline U_1\cap\dots\cap\overline U_n$ is not empty.

Now take any disjoint nonempty open sets $U,V$ in $X$. Assuming that $V$ is of type $F_\sigma$, we can apply the Baire Theorem and find a nonempty open set $W\subseteq V$ whose closure in $X$ is contained in $V$. Then $\overline{U}\cap\overline{W}=\emptyset$, which contradicts the superconnectedness of $X$. $\square$

The cardinality $\mathrm{non}(\mathcal M)$ is the above example can be lowered to $\mathfrak q_0$, where $\mathfrak q_0$ is the smallest cardinality of a second-countable metrizable space which is not a $Q$-space (= contains a subset which is not of type $G_\delta$).

A topological space is submetrizable it it admits a continuous metric. Each submetrizable space is functionally Hausdorff in the sense that for any distinct elements $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ such that $f(x)\ne f(y)$.

Example 2. There exists a submetrizable second-countable space $X$ of cardinality $\mathfrak q_0$, which is not symmetrizable.

Proof. By the definition of the cardinal $\mathfrak q_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap_{n\in\omega}W_n$ of some open sets $W_n\in\tau'$. By the choice of the topology $\tau'$, for every $n\in\omega$ there exists open sets $U_n,V_n\in \tau$ such that $W_n=U_n\cup(V_n\setminus A)$. It follows from $A\subseteq W_n=U_n\cup(V_n\setminus A)$ that $A=A\cap W_n=A\cap U_n\subseteq U_n$. $$A=\bigcap_{n\in\omega}W_n=A\cap\bigcap_{n\in\omega}W_n=\bigcap_{n\in\omega}(A\cap W_n)=\bigcap_{n\in\omega}(A\cap U_n)\subseteq \bigcap_{n\in\omega}U_n\subseteq \bigcap_{n\in\omega}W_n=A$$ and hence $A=\bigcap_{n\in\omega}U_n$ is a $G_\delta$-set in $X$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$

On the other hand we have the following partial affirmative answer to Question 1.

Theorem 3. Martin's Axiom implies that every second-countable $T_1$ space of cardinality $<\mathfrak c$ is perfect and hence symmetrizable.

Proof. It is known that Martin's Axiom implies that every second-countable $T_1$-space $X$ of cardinality $\mathfrak c$ is a $Q$-space, which means that every subset of $X$ is of type $G_\delta$. In particular, $X$ is perfect and by Theorem 2 is symmetrizable. $\square$

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    $\begingroup$ For the record, I didn't know that either or both theorems were consequences of a more general result in Arhangelski's paper -- I didn't look at the paper much, other than to see that the symmetrizable property appears non-trivially in the paper. More accurate would be to say something like: "After looking at a reference Dave L. Renfro suggested in a comment as possibly being relevant, I discovered that both theorems follow from the known ..." $\endgroup$ May 30 at 21:02
  • $\begingroup$ @DaveLRenfro Thank you for the suggestion of editing my answer. What is surprising that reading the proof of Arhangelski I found that it works only for Hausdorff spaces. So, the case of $T_1$ spaces seems to be open. $\endgroup$ May 30 at 21:53
  • $\begingroup$ Correct me if I'm wrong, but symmetrisability and semimetrisability are equivalent in Fréchet-Urysohn spaces, while every semimetrisable space with a point-countable base is developable (I don't thing anything here requires regularity, but I didn't check). B. Scott gives here an example of a second-countable $T_2$ space which is not developable. By the above, it cannot be symmetrisable. $\endgroup$
    – Tyrone
    May 31 at 12:50
  • $\begingroup$ @Tyrone Thank you for your comment. In fact, besides your or my example there are even more simple examples of submetrizable second-countable submetizable spaces which are not symmetrizable. Now I will add a suitable example to my answer. $\endgroup$ May 31 at 13:02

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