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Let $X$ be a complete metric space. Show that $E \subset X$ is nowhere dense if and only if for each open subset $U$ of $X$, the intersection $E \cap U$ is not dense in $U$. We say that $E$ is nowhere dense if $\text{int}(\overline{E}) = \emptyset$.

My attempt:

($\Rightarrow$) Assume $E \subset X$ is nowhere dense. Suppose for the sake of contradiction that $E \cap U$ is dense in $U$ for some open set $U$. Then $\overline{E \cap U} = U$, and so $U = \overline{E \cap U} \subset \overline{E}$. But because $E$ is nowhere dense, $\overline{E}$ contains no open sets. This is a contradiction, and so we must have that $E \cap U$ is not dense in $U$. \

($\Leftarrow$) Assume that for each open subset $U$ of $X$, the intersection $E \cap U$ is not dense in $U$. Suppose for the sake of contradiction that $E$ is not nowhere dense. This means that for some nonempty open set $U$, $\text{int}(\overline{E}) = U$. I feel like this should lead to a contradiction, but I cannot quite see how it falls out.

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  • $\begingroup$ The terms open, interior, closed, closure, dense, nowhere dense are all relative concepts -- relative to a containing space (which is $X$ by default unless either specified otherwise or when the context implies a different containing space), Thus, the unqualified statement "$E$ is nowhere dense" means $E$ is nowhere dense in $X$. $E$ is automatically dense in itself. If $E$ is nonempty, $E$ has nonempty interior in $E$, even though $E$ has empty interior in $X$. $\endgroup$ – quasi Jan 27 '17 at 7:42
  • $\begingroup$ Also, are you sure the problem is correctly stated? For example, let $U = X$. Then if $E$ is nonempty, $E \cap U = E$ which is dense in $E$. $\endgroup$ – quasi Jan 27 '17 at 7:48
  • $\begingroup$ I am not sure if it is true that $E$ is always dense in itself. $\endgroup$ – Ann Marie Jan 28 '17 at 19:25
  • $\begingroup$ Right, my mistake. I meant to say: "$E$ is automatically dense in $E$" (since $E$ closure is $E$). Note that the question asked if $E \cap U$ is always dense in $E$, and if I understand the definitions correctly, the answer is trivially "no" for the case $U = X$. $\endgroup$ – quasi Jan 28 '17 at 21:17
  • $\begingroup$ I am still not sure, why should $E$ closure be always $E$? This is only true if $E$ is closed, but we don't have anything that says this is the case. To my understanding, nowhere dense sets need not be closed. $\endgroup$ – Ann Marie Jan 28 '17 at 21:27
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(1) . You should refer to only non-empty open $U$ in both parts .

(2).($\implies$) To be consistent, as you already used the closure bar to denote closure in $X$, you should say $E\cap U$ is dense in $U$ iff $\overline {E\cap U}\supset U.$ (Not $\overline {E\cap U}=U.$) Otherwise you are correct, except to say that $\overline E$ has no non-empty open subsets.

(3). If $E$ is nowhere dense and $U$ is open and not empty then $E$ cannot be dense in $U,$ otherwise we have $\overline {E\cap U}\supset U$, implying $$\phi =int(\overline E)\supset int (\overline {E\cap U})\supset int (U)=U\ne \phi$$ a contradiction.

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