Metrizable topology given by convergence of sequences

Question

Let $\alpha$ be a (Hausdorff) topology on a set $X$. Let $T_{\alpha} = \{ \beta \text{ topology on } X \mid \forall (x_n)_{n \in \mathbb{N}} \text{ in } X, x \in X: x_n \overset{\alpha} {\longrightarrow} x \Longleftrightarrow x_n \overset{\beta}{\longrightarrow} x \}$ be the set of all topologies $\beta$ on $X$ that agree on sequences with $\alpha$.

The question is: Does there exist any metrizable topology in $T_{\alpha}$? If so, is it unique?

Answer 1

For topologies $\alpha, \beta$ on $X$ define:

1. $\alpha \rightarrow \beta$, if for all sequences $(x_n)_{n \in \mathbb N}$ and $x \in X$ it holds: $(x_n)_{n \in \mathbb N} \overset{\alpha} {\longrightarrow} x \Longrightarrow (x_n)_{n \in \mathbb N} \overset{\beta}{\longrightarrow} x $
2. $\alpha \leftrightarrow \beta$, if $\alpha \rightarrow \beta$ and $\beta \rightarrow \alpha$

Obviously, $\rightarrow$ is reflexive and transitive on the set of all topologies on $X$, hence $\leftrightarrow$ is an equivalence relation and $T_\alpha$ as defined above is the equivalence class of $\alpha$.

Recall that a topological space is sequential, if for each non-closed set $A$ there exists a sequence in $A$, which converges to an element outside of $A$. Obviously, each first countable space is sequential.

The following is easy to see:

Let $\alpha, \beta$ be topologies on $X$.

1. $\beta \subset \alpha \Rightarrow \alpha \rightarrow \beta$
2. If $\alpha$ is sequential, then $\beta \subset \alpha \Leftrightarrow \alpha \rightarrow \beta$
3. If $\alpha, \beta$ are sequential, then $\beta = \alpha \Leftrightarrow \alpha \leftrightarrow \beta$

Hence:
If $\alpha$ is sequential and $T_\alpha$ as defined above contains a metrizable topology, then $\alpha$ is metrizable.

This provides a lot of topologies $\alpha$, such that $T_\alpha$ contains no metrizable topology.