# Is each cosmic space cometrizable?

A regular topological space $$X$$ is called

$$\bullet$$ cosmic if $$X$$ is a continuous image of a separable metrizable space;

$$\bullet$$ cometrizable if $$X$$ admits a weaker metrizable topology such that each point $$x\in X$$ has a (not necessarily open) neighborhood base consisting of metrically closed sets.

In 1989 Gruenhage proved that under PFA a cometrizable space is cosmic if and only if it contains no uncountable discrete subspace and no uncountable subspace of the Sorgenfrey line.

Can the cometrizability be moved to the right-hand part of this characterziation?

Question. Is each cosmic space cometrizable? Maybe under PFA or OCA?

• Is there a promising candidate for the metric? I mean, given a quotient of a metric space, if it is not too bad one can try the distance between points to be minimum distance between their preimages, say. Is it known that such simplistic attempts are hopeless? – მამუკა ჯიბლაძე Jan 22 at 19:14
• If such metric exists, then any stronger continuous metric also makes the job. So either a metric witnessing the submetrizability exists and then there are plenty of them or it does not exist at all. – Taras Banakh Jan 22 at 20:59
• Are there cometrizable spaces without the weakest metric that does the job? – მამუკა ჯიბლაძე Jan 23 at 5:29
• @მამუკაჯიბლაძე The space $\mathbb Q$ of rational numbers is (co)metrizable but does not have a minimal cometrizing metrizable topology. – Taras Banakh Jan 23 at 6:49
• Are you asking if the cosmos is a comet space? – Asaf Karagila Jan 24 at 8:16

Discussing this problem with Alex Ravsky we constructed the following

Example. The Euclidean topology $$\tau_0$$ on the set $$\mathbb Q$$ of rational numbers can be enlarged to a regular topology $$\tau$$ of weight $$\omega_1$$ such that the countable (and hence cosmic) topological space $$(\mathbb Q,\tau)$$ is not cometrizable.

The topology $$\tau$$ is constructed by transfinite induction of length $$\omega_1$$. At each step $$\alpha<\omega_1$$ we define a regular second countable topology $$\tau_\alpha$$ on $$\mathbb Q$$ such $$\bigcup_{\beta<\alpha}\tau_\beta\subset\tau_\alpha\subsetneq \tau_{\alpha+1}$$ that the space $$(\mathbb Q,\tau_\alpha)$$ has no isolated points and for every $$\alpha<\omega_1$$ the topology $$\tau_{\alpha+1}$$ contains a neighborhood $$U_\alpha$$ of zero such that for every $$\beta\ge\alpha$$ and any neighborhood $$V\in\tau_\beta$$ of zero the $$\tau_\alpha$$-closure of $$V$$ is not contained in $$U_\alpha$$. Then the topology $$\tau=\bigcup_{\alpha\in\omega_1}\tau_\alpha$$ is a desired regular topology for which the space $$(\mathbb Q,\tau)$$ is not cometrizable. $$\square$$

On the other hand, we have many positive results, see this preprint for more information.

For topological spaces $$X,Y$$ by $$C_k(X,Y)$$ we denote the space of continuous functions from $$X$$ to $$Y$$, endowed with the compact-open topology.

Theorem 1. For any $$\aleph_0$$-space $$X$$ and any cometrizable space $$Y$$ the function space $$C_k(X,Y)$$ is cometrizable.

Proof. It is well-known that the $$\aleph_0$$-spaces are images of metrizable separable spaces under compact-covering maps, which implies that $$C_k(X,Y)$$ embeds into the function space $$C_k(M,Y)$$ over some metrizable separable space $$M$$. So, we can assume that the space $$X$$ is metrizable and separable. Let $$D$$ be a countable dense set in $$X$$. Let $$\tau$$ be a metrizable topology witnessing that the space $$Y$$ is cometrizable. It can be shown that the topology on $$C_k(X,Y)$$ inherited from the Tychonoff power $$(Y,\tau)^D$$ witnesses that the function space $$C_k(X,Y)$$ is cometrizable.

Corollary 1. For any $$\aleph_0$$-space $$X$$ the function spaces $$C_k(X)=C_k(X,Y)$$ and $$C_k(C_k(X))$$ are cometrizable.

A Tychonoff space $$X$$ is Ascoli if the canonical map $$\delta:X\to C_k(C_k(X))$$ assigning to each $$x\in X$$ the Dirac measure $$\delta_x:f\mapsto f(x)$$ is a topological embedding. It is known that each $$k$$-space is Ascoli.

The definition of an Ascoli space and Corollary 1 implies

Corollary 2. Each Ascoli $$\aleph_0$$-space is cometrizable. In particular, each sequential $$\aleph_0$$-space is cometrizable.

Finally, we have

Theorem 2. Each stratifiable space $$X$$ is cometrizable.

Proof. Since $$X$$ is stratifiable, every point $$x\in X$$ has a decreasing system of open neighborhoods $$(W_n(x))_{n\in\omega}$$ such that each closed set $$F\subset X$$ is equal to $$\bigcap_{n\in\omega}\overline{W_n[F]}$$ where $$W_n[F]=\bigcup_{x\in F}W_n(x)$$.

It is known that each stratifiable space is a $$\sigma$$-space. So $$X$$ has a $$\sigma$$-discrete network $$\mathcal N$$, which can be written as the countable union $$\mathcal N=\bigcup_{k\in\omega}\mathcal N_k$$ of discrete families in $$X$$. Since stratifiable spaces are paracompact, each set $$N\in\mathcal N_k$$ has an open neighborhood $$O_N\subset X$$ such that the family $$(O_N)_{N\in\mathcal N_k}$$ is discrete in $$X$$.

For every $$k,n\in\omega$$ and $$N\in\mathcal N_k$$ consider the open neighborhood $$O_{k,N,n}=O_N\cap W_n[N]$$ of $$N$$. Since stratifiable spaces are perfectly normal, there exists a continuous function $$f_{k,N,n}:X\to [0,1]$$ such that $$f_{k,N,n}^{-1}(1)=N$$ and $$f_{k,N,n}^{-1}(0)=X\setminus O_{k,N,n}$$.

Let $$\ell_1(\mathcal N_k)$$ be the Banach space of all functions $$h:\mathcal N_k\to\mathbb R$$ such that $$\|h\|:=\sum_{N\in\mathcal N_k}|h(N)|<+\infty$$. For every $$N\in\mathcal N_k$$ let $$e_N:\mathcal N_k\to\{0,1\}$$ be the unique function such that $$e_N^{-1}(1)=\{N\}$$. So $$(e_N)_{N\in\mathcal N_k}$$ is the standard unit basis of the Banach space $$\ell_1(\mathcal N_k)$$.

Consider the continuous function $$f_{k,n}:X\to\ell_1(\mathcal N_k),\;\;f_{k,n}:x\mapsto\sum_{N\in\mathcal N_k}f_{k,N,n}(x)e_N.$$ Next, consider the continuous (injective) function $$f:X\to\prod_{k\in\omega}\ell_1(\mathcal N_k)^\omega,\;\;f:x\mapsto \big((f_{k,n}(x))_{n\in\omega}\big)_{k\in\omega}.$$

Let $$\tau$$ be the metrizable topology on $$X$$ such that the function $$f:(X,\tau)\to\prod_{k\in\omega}\ell_1(\mathcal N_k)^\omega$$ is a topological embedding. It follows that for any $$k,n\in\omega$$ and $$N\in\mathcal N_k$$ the set $$N$$ is $$\tau$$-closed and the set $$O_{k,N,n}$$ is $$\tau$$-open.

We claim that the topology $$\tau$$ witnesses that the space $$X$$ is cometrizable. Given any point $$x\in X$$ and an open neighborhood $$O_x\subset X$$ of $$x$$, use the regularity of $$X$$ to find an open neighborhood $$U$$ of $$x$$ such that $$\overline{U}\subset O_x$$. Consider the closed set $$F=X\setminus U$$ and observe that $$F=\bigcap_{n\in\omega}\overline{W_n[F]}$$. Since $$x\notin F$$, there exists $$n\in\omega$$ such that $$x\notin\overline{W_n[F]}$$. Then $$V_x:=X\setminus\overline{W_n[F]}$$ is an open neighborhood of $$x$$.

Since $$\mathcal N$$ is a network, the open set $$X\setminus\overline{U}\subset F$$ coincides with the union $$\bigcup\mathcal N'$$ of the subfamily $$\mathcal N'=\{N\in\mathcal N:N\subset X\setminus\overline{U}\}$$. For every $$k\in\omega$$ let $$\mathcal N_k'=\mathcal N_k\cap\mathcal N'$$. Observe that the union $$W=\bigcup_{k\in\omega}\bigcup_{N\in\mathcal N'_k}O_{k,N,n}$$ is a $$\tau$$-open set such that $$X\setminus\overline{U}\subset W=\bigcup_{k\in\omega}\bigcup_{N\in\mathcal N'_k}O_{k,N,n}\subset W_n[X\setminus\overline{U}]\subset W_n[F]\subset X\setminus V_x.$$ Then $$V_x\subset X\setminus W\subset \overline{U}\subset O_x$$ and the $$\tau$$-closure of the neighborhood $$V_x$$ is contained in $$X\setminus W\subset O_x$$. $$\square$$