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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?

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    $\begingroup$ 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 $B(a,r)=A$. $\endgroup$ Aug 11, 2021 at 11:28
  • $\begingroup$ 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. $\endgroup$ Aug 11, 2021 at 11:31
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    $\begingroup$ Your space is, in fact, metrizable! $\endgroup$ Aug 11, 2021 at 11:52
  • $\begingroup$ @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... $\endgroup$ Aug 11, 2021 at 21:14

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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.

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    $\begingroup$ $(X,J)$ is still metrizable by a pseudometric. $\endgroup$
    – Magma
    Aug 11, 2021 at 15:54
  • $\begingroup$ @Magma I know but that wasn’t the question. $\endgroup$ Aug 11, 2021 at 16:10

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