Is the space of metric topologies over a given set dense (in the order sense)? 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?
 A: $\def\cl{\operatorname{cl}}$
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.
