I have this question:

Let $(X, d_1)$ and $(X,d_2)$ be two metric spaces. Suppose $d_1$ is topologically finer than $d_2$. What is the relationship between these two statements?

(i) $(X,d_1)$ is separable.

(ii) $(X,d_2)$ is separable.

I think the implication (i) $ \Rightarrow$ (ii) holds, but not the other way around. I was trying to prove it:

Suppose $(X,d_1)$ is separable. Then $X$ has an at most countable dense subset for the $d_1$ metric. So there exists an at most countable subset $A \subset X$ such that $\overline{A} = X$. Since $d_1$ is topologically finer than $d_2$, every open subset for the $d_2$ metric is also open for the $d_1$ metric. This is equivalent in saying that every closed $d_2$-subset is also $d_1$-closed.

This is as far I could go. What can I conclude from this? And am I also correct in saying the other implication is false?

  • $\begingroup$ Sorry, I misread. You are correct is thinking that separability in a finer topology implies separability in a coarser topology. I have deleted temporarily my answer and will repair. That's done now. I hope I did not cause any confusion. $\endgroup$ – drhab May 17 '16 at 9:30
  • $\begingroup$ I saw your answer, and I was thinking you confused $d_1$ with $d_2$. But your answer still helped me a lot in giving me the right idea! $\endgroup$ – Kamil May 17 '16 at 9:36
  • $\begingroup$ Nice to hear that. There was indeed some confusion by me about $d_1,d_2$. $\endgroup$ – drhab May 17 '16 at 9:38

Counterexample of (ii)$\implies$(i):

Let $d_1$ be defined on $\mathbb R$ by $d_1(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise.

Let $d_2$ be defined on $\mathbb R$ by $d_2(x,y)=|x-y|$.

Then $\mathbb R$ equipped with $d_1$ results in the discrete topology (finest as possible), hence no separability on the uncountable $\mathbb R$.

On the other hand $\mathbb R$ equipped with $d_2$ results in the usual topology, hence separability.

Proof of (i)$\implies$(ii):

Let $X$ equipped with some topology $\tau$ be separable.

Then a countable and dense $S\subseteq X$ exists.

Density of $S$ means exactly that $U\cap S\neq\varnothing$ for each $U\in\tau$.

If $\sigma$ is another topology on $X$ and $\tau$ is finer then also $U\cap S\neq\varnothing$ for $U\in\sigma$. This because $\sigma\subseteq\tau$.

So again $S$ is dense. Of course it has not lost its countability and we conclude that $X$ equipped with the coarser $\sigma$ is separable.


A counterexample for the opposite: We use the set $\Bbb R$ with usual metric $d_2(x,y)=|x-y|$ and define $$d_{1}\left(x,y\right)=\cases{ 0&\text{if }x=y\cr 1&\text{if }x\neq y}$$ We know that $d_2$ make a separable space (use $\Bbb Q$). $d_1$, on the other hand, yields discrete topology, so the only dense set is the whole space, making it not separable.

$d_1$'s topology is finer than $d_2$'s as it's discrete and therefore finer than any other topology.

  • $\begingroup$ $ x \not = y$ \not = $\endgroup$ – Olba12 May 17 '16 at 8:50
  • $\begingroup$ Ah, I see thanks. But is $d_1$ topologically finer than $d_2$ ? $\endgroup$ – Kamil May 17 '16 at 8:54
  • $\begingroup$ @Kamil Every set open in $d_2$ (in fact every set) is open in $d_1$. $\endgroup$ – Heimdall May 17 '16 at 8:59
  • $\begingroup$ @Olba12 When I was typing the answer, I tried different combinations: \ne, \neq, \not=, \not =; the preview didn't render either of them correctly. This question uses \neq, I'm quoting the source: If det $ A\neq0$, the matrix, so what's wrong with \neq and other possible no-equals in my answer? Maybe it works in question and comments but not in answers? I'm asking on meta. $\endgroup$ – Heimdall May 17 '16 at 9:46
  • $\begingroup$ Weird, \not = has always been working for me... haha $\endgroup$ – Olba12 May 17 '16 at 10:08

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