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Prove that a set $A$ is dense in a metric space $(M,d)$ iff $A^c$ has empty interior.

Attempt:

I think I proved the converse correctly, but I'm not sure how to start the forward direction.

$(\Leftarrow)$ Assume $A^c$ has empty interior. Show that $M=\operatorname{cl}(A)$, where $\operatorname{cl}(A)$ denotes the closure of $A$. We also note that $\operatorname{cl}(A)=(\operatorname{int}(A^c))^c=\emptyset^c=M$, where $\operatorname{int}(A^c)$ denotes the set of interior points of $A^c$.

$\blacksquare$

Any help would be appreciated. Thanks.

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This stems from the more general fact (see this question) that given $A \subseteq X$ $$\operatorname{Int} (A) = X \setminus \overline{ X \setminus A },$$ or, equivalently, $\operatorname{Int}( X \setminus A ) = X \setminus \overline{A}$. (Note that this is valid in all topological spaces, not just metric spaces.)

So $A$ is dense if and only if $X = \overline{A}$ if and only if $\varnothing = X \setminus X = X \setminus \overline{A} = \operatorname{Int} ( X \setminus A )$.

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$\implies$

Suppose that $A$ is dense in $M$. Then $cl(A)=M$. If $M$ is empty then the result is trivial. So assume $M\ne\phi$. The $cl(A)\ne \phi$. Then clearly, $A\ne \phi$ otherwise $A=\phi=cl(\phi)\ne M$. Thus $A\ne \phi \implies A^c=\phi$ and so its interior. $\blacksquare$

Converse what you have proved seems to be right.

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  • $\begingroup$ What if $A = \mathbb{Q}$ and $M = \mathbb{R}$? $\endgroup$ – fixedp Sep 19 '14 at 8:27
  • $\begingroup$ $\mathbb{Q}$ is dense in $\mathbb{R}$. Also the set of irrationals has empty interior. $\endgroup$ – Hirak Sep 19 '14 at 8:33
  • $\begingroup$ True, I suggest you take another look at the last implication in your proof. $\endgroup$ – fixedp Sep 19 '14 at 8:45
  • $\begingroup$ @fixedp, Correct !! I will rectify that soon ! Thanks $\endgroup$ – Hirak Sep 20 '14 at 7:15
  • $\begingroup$ why $A \neq \emptyset \Rightarrow A^{c} = \emptyset$? $\endgroup$ – user286485 Oct 11 '16 at 20:46
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Just recall that in any topological space $X$ the closure of the complement is equal to the complement of the interior. Formally $(E^\circ)^c=\overline{(E^c)}$ for any $E\subseteq X$. Take $E^\circ=\emptyset$ for one implication. Assume $E^c$ to be dense to get the other implication.

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