Is every topology metric inducible?

So I'm trying to show that not every topology is metric inducible.

I take a set $$X$$. Now take $$J = \{\emptyset, X, A, X-A\}$$ where $$A \subset X$$.

Assume the metric space $$(X,d)$$ induces $$J$$. Now since $$A$$ is open in $$X$$, for $$a \in A$$ there exists some $$r > 0$$ such that $$B(a,r) \subset A$$. But the only open sets in $$(X,d)$$ are $$\emptyset, X, A, X-A$$. This implies $$B(a,r) = \emptyset$$ which isn't possible.

So we conclude $$J$$ isn't inducible by any metric.

Is this proof okay?

• It is not correct. You cannot say there exist $r$ such that $B(a,r)$v is a proper subset of $A$. We may have $B(a,r)=A$. Aug 11, 2021 at 11:28
• If I understand you right, you're trying to find a topology that is not metrizable--that there is no metric yielding that topology. You might look at the finite complement topology example. Aug 11, 2021 at 11:31
• Your space is, in fact, metrizable! Aug 11, 2021 at 11:52
• @KaviRamaMurthy You mean pseudo-metrisable, by $d(x,y)=0$ if $\{x,y\} \subseteq A$ or $\{x,y\} \cap A=\emptyset$ and $d(x,y)=1$ otherwise... Aug 11, 2021 at 21:14

That proof cannot be correct: if $$X=\{1,2\}$$ and $$A=\{1\}$$ in fact we do have a metrisable space (a discrete two point space). Note that we cannot claim that $$B(a,r) \subset A$$ but only $$B(a,r) \subseteq A$$; the inclusion need not be proper, as you seem to think.
But if $$|A| \ge 2$$ or $$|X-A|\ge 2$$ the topological space $$(X,J)$$ is indeed non-metrisable as there is no open set "separating" two points in that one open set, so $$(X,J)$$ is not $$T_0$$ while all metric-induced topologies are.
• $(X,J)$ is still metrizable by a pseudometric. Aug 11, 2021 at 15:54