# 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?

• I think you also want the "in the metric sense" 'dense', in addition to the "in the order sense" 'dense', in your title. (And, of course, infinitude of $S$ is redundant.) Commented Sep 12, 2020 at 0:53

$$\def\cl{\operatorname{cl}}$$ 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.

• Great! Thanks very much. Commented Oct 1, 2020 at 15:07