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?*




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**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!



Is every second-countable Hausdorff space symmetrizable?
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
$$d(x,y)=\inf\big\{2^{-n}:\max\{x,y\}\in U\_n(\min\{x,y\})\big\}.\quad\square$$




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**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](http://www.mathnet.ru/links/dbf92f01dafb7bfdaa4d7d96dd652ad7/rm5901.pdf) (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.




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**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"](https://www.sciencedirect.com/science/article/pii/B9780444865809500136).


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](https://www.ams.org/journals/proc/1980-078-02/S0002-9939-1980-0550513-4/S0002-9939-1980-0550513-4.pdf) 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. $\square$


