proposition: Let $(X,d_X)$ be a complete metric space. $A$ is nowhere dense if and only if for every open subset $O\subset X$ there is a open ball $B(x,\epsilon)$ (contained in $O$) such that $A\cap \text{cl}(B(x,\epsilon))=\emptyset$

How can I prove this proposition? I know that a subset $A\subset X$ is nowhere dense if $\text{int}(\text{cl}(A))=\emptyset$, but I have no idea how to begin. Any tips/advice? Thx

  • $\begingroup$ Note that the open subset $O \subset X$ must be nonempty. $\endgroup$ – user642796 Sep 23 '13 at 12:32
  • 2
    $\begingroup$ Is $O$ supposed to have some role in the proposition? $\endgroup$ – Clayton Sep 23 '13 at 12:33
  • $\begingroup$ @ArthurFischer why is that? $\endgroup$ – jack Sep 23 '13 at 13:16
  • $\begingroup$ @jack: There are no balls contained in $\varnothing$ (unless you count the empty ball as a ball, but then the empty ball is a subset of every open set, meaning that -- apparently -- all sets are nowhere dense according to this equivalence). $\endgroup$ – user642796 Sep 23 '13 at 14:16

HINT: Suppose first that for each non-empty open set $U\subseteq X$ there is an open ball $B(x,\epsilon)\subseteq U$ such that $A\cap\operatorname{cl}B(x,\epsilon)=\varnothing$; you want to show that $\operatorname{int}\operatorname{cl}A=\varnothing$. Let $U=\operatorname{int}\operatorname{cl}A$; certainly $U$ is open. If $U\ne\varnothing$, there is an open ball $B(x,\epsilon)\subseteq U$ such that $A\cap\operatorname{cl}B(x,\epsilon)=\varnothing$. Clearly $A\cap B(x,\epsilon)=\varnothing$; combine this with the fact that $x\in U$ to get a contradiction.

Now suppose that $A$ is nowhere dense in $X$, so that $\operatorname{int}\operatorname{cl}A=\varnothing$, and let $U$ be a non-empty open set in $X$; you want to show that there are an $x\in U$ and an $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$ and $A\cap\operatorname{cl}B(x,\epsilon)=\varnothing$. $U\nsubseteq\operatorname{cl}A$ (why?), so pick any $x\in U\setminus\operatorname{cl}A$. There is a $\delta>0$ such that $B(x,\delta)\cap A=\varnothing$ (why?). Now how can you choose $\epsilon>0$ so that $A\cap\operatorname{cl}B(x,\epsilon)=\varnothing$ and $B(x,\epsilon)\subseteq U$?


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