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Suppose $X$ is a metric space. Take $F \subset X, F \neq \emptyset$ such that $int\;F = \emptyset$. I can clearly see that $F^c$ is dense in $X$ if $X$ is a Euclidean space. But unable to see it for a general metric space. Any help is much appreciated.

Canine360
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  • Can you give a proof how this works in Euclidean space? – Severin Schraven Jul 25 '21 at 03:10
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    Well the boundary of F is its closure minus interior. interior is empty, so boundary =closure =F because F is closed. Suppose, BWOC, there exists an open set such that its intersection with the complement of F is empty. Which means it is a subset of F, therefore a subset of the boundary of F. But that is not possible because the boundary of F has an empty interior. – Canine360 Jul 25 '21 at 03:20
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    Your proof works in any space and is already better than mine. I would also add that the boundary consideration is not necessary. If there is an open set $U$ in $X\setminus F$, then $X\subseteq F$, so $F$ has nonempty interior. – pancini Jul 25 '21 at 03:21
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    Thanks so much! Of course! I see it now. If the intersection of the open set with the complement of F is empty, it must be a subset of F, which is a contradiction. – Canine360 Jul 25 '21 at 03:29
  • @Canine360 What theorem are you using to assert that the boundary of $F$ has empty interior? This isn't true for a general set $F$ (consider $F = \mathbb Q$ and $X = \mathbb R$). If you're asserting that it's true when $F$ is closed, isn't that essentially equivalent to what you're trying to prove? Seems your argument could be circular, but it depends on how you proved that fact you are using. –  Jul 25 '21 at 03:33
  • First, I considered it well known that the boundary of an open or closed set is empty, see here, e.g. https://math.stackexchange.com/questions/1142892/prove-that-the-interior-of-the-boundary-is-empty. It does not seem circular because the proof of emptyness of the int of the bd linked above has got nothing to do with denseness. Secondly you can forget about the boundary, see Elliot's comment above which simplifies my proof further. – Canine360 Jul 25 '21 at 11:14

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Take $x\in X$ and let $r=\inf\{d(x,y):y\in X\setminus F\}$. It suffices to show $r=0$.

If $r>0$, then $\emptyset\neq B_r(x)\subseteq F$.

Edit: see OPs comment for a better answer.

pancini
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