# Is every second-countable Hausdorff space symmetrizable?

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.

• "A symmetric"? this sounds (to me — not native) as phoney English.
– YCor
May 30 at 17:55
• @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. May 30 at 18:01
• 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. May 30 at 19:40
• @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. May 30 at 19:59
• @Ycor Perhaps it's a pun on the word "metric", so one should read it as "a sym-metric"? Jun 1 at 8:38

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. $$\square$$

• 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 ..." May 30 at 21:02
• @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. May 30 at 21:53
• 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. May 31 at 12:50
• @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. May 31 at 13:02