Suppose that $S$ is an infinite set and that $\alpha$ and $\beta$ are metrics over $S$ such that the topology induced by $\alpha$ is everywhere strictly finer than the metric induced by $\beta$, meaning that every open set $U$ in $\beta$ contains a set $V$ that is open in $\alpha$ but not in $\beta$. Suppose further that $S$ is dense (in the metric sense) with respect to both metrics, in the sense that, for any $x \in S$, for any $\epsilon > 0$ there exists $y \neq x$ within distance $\epsilon$ of $x$. Does there exist a metric $\gamma$ over $S$ that is everywhere strictly finer than $\beta$ but everywhere strictly coarser than $\alpha$? Or, contrarily, are there cases where it is known that no such $\gamma$ exists?



Is the space of metric topologies over a given set dense (in the order sense)?
$\def\cl{\operatorname{cl}}$
A large family of counterexamples can be constructed using the following
proposition:


Let $(S, T\_1)$ be a topological space with two complementary dense
subspaces $A, B$.
Define $T\_3 = \{ (A \cap U) \cup (B \cap V) \mid U, V \in T\_1 \}$,
in other words $(S, T\_3)$ is the topological sum of $A$ and $B$.
Let $T\_2$ be a topology on $S$, finer than $T\_1$, such that $T\_3$ is
everywhere strictly finer than $T\_2$.
If $T\_2$ is semi-regular, then $T\_2 = T\_1$.


Sketch of a proof:


* Observe that $T\_1|A = T\_2|A = T\_3|A$ and $T\_1|B = T\_2|B = T\_3|B$. Moreover,
$\cl\_3 V = \cl\_A (V \cap A) \cup \cl\_B (V \cap B)$ for every $V \subset S$.
* From the assumption that $T\_3$ is everywhere strictly finer than $T\_2$
it follows that both $A$ and $B$ are dense in $(S, T\_2)$. Hence for
every $U \in T\_2$ we have $\cl\_2 (U \cap A) = \cl\_2 (U \cap B) = \cl\_2 U$.
* Show that for any $U \in T\_2$ we have $\cl\_2 U = \cl\_3 U = \cl\_1 U$.
* Deduce that if $U$ is a regular open set in $(S, T\_2)$, then $U \in T\_1$.
Therefore, if $T\_2$ is semi-regular, $T\_2 \subset T\_1$.


The application to your problem is easy. If $(S, T\_1)$ is nonempty,
metrizable and dense in itself, there are many choices of $A,B$ and
$T\_3$ will also be metrizable and dense in itself. Of course for any
$U \in T\_1 \setminus \{\emptyset\}$ we have $U \cap A \in T\_3 \setminus T\_1$
and if $T\_2$ is to be metrizable it must certainly be semi-regular.


