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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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    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 $X$. – 2017-01-27
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    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 $E$. – 2017-01-27
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    I am not sure if it is true that $E$ is always dense in itself. – 2017-01-28
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    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 $U = X$. – 2017-01-28
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    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. – 2017-01-28
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    $E$ is closed in $E$. $E$ need not be closed in $X$. – 2017-01-28
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    But, I think I've added confusion. I think my objections are based on a misinterpretation of the terminology in the context of this problem. If the phrase "$E \cap U$ is dense in $E$" was intended to be interpreted as "$E \cap U$ is dense in itself", then the problem is fine, and my comments should ignored -- sorry. – 2017-01-28
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    I haven't re-examined the problem based on the "dense in itself" interpretation, but one question -- where have you used the hypothesis that $X$ is a complete metric space? For the direction you did, wouldn't the same argument work in any topological space? Perhaps the the condition "$X$ is a complete metric space" is superfluous, or maybe it's only needed for the converse. Just asking. – 2017-01-28
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    I'm a little rusty, but now that I understand the problem ... – 2017-01-28
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    Your argument for the forward direction is correct, but there's no need for a proof by contradiction -- there's no benefit in terms of bervity or clarity. – 2017-01-28
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    Forgive my confusion -- I used to be quite good at this stuff, but as I said, I'm a little rusty. But looking at your argument again, I'm not sure it works. You proved $E \cap U$ is nowhere dense. You then claim that implies $E \cap U$ is not dense in $E$. But consider the Cantor set. It's nowhere dense, but it's dense in itself. – 2017-01-28
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    So let $X = \mathbb{R}$, $E = \text{the Cantor set}$, $U=X$. Then $E$ is nowhere dense (in $X$), but $E \cap U = E$, which is dense in $E$ (since the Cantor set is dense in itself. Doesn't that mean that the forward implication of the problem is wrong? – 2017-01-28
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    I think you may be correct, but I am pretty confused now...I am just going to wait until Monday and ask my prof about it. Thank you! – 2017-01-28
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    OK, and I apologize twice -- (1) for not helping; (2) for my adding my own confusion to yours. – 2017-01-28
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    But just to clarify two definitions . . . Let $E \subseteq X$, where $X$ is a topological space. Definition of nowhere dense: $E$ is nowhere dense if int $\overline{S}=\emptyset$. Definition of dense in itself: $E$ is dense in itself if $E$ contains no isolated points. Agreed? – 2017-01-28
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    And one more definition (just to get the terminology straight) ... Let $S \subseteq E \subseteq X$, where $X$ is a topological space. Definition of "$S$ is dense in $E$": $S$ is dense in $E$ if $\overline S=\overline E$. Yes? If so, the statement "$E$ is dense in $E$ is automatically true, while the statement "$E$ is dense in itself" may or may not be true. – 2017-01-29
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    Turns out there was a typo. It is supposed to be that the intersection is not dense in $U$, not $E$. I just fixed it. – 2017-01-30
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    Right -- I was pretty sure the problem as originally posted was misstated. The new statement makes sense, and seems a routine verification. Just apply the definitions. – 2017-01-30
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    A minor issue: I think the statement should be "... for each nonempty open subset $U$ of $X$ ..." – 2017-01-30
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    One more concern: As far as I can see, the proof of the revised statement only needs a topological argument. Thus, unless I'm missing something, the specification "Let $X$ be a complete metric space" could be weakened to "Let $X$ be a topological space". Of course, I could be missing something. – 2017-01-30
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    I guess your definition of "$E$ is nowhere dense" is ***not*** "there is no nonempty open set $U$ such that $E\cap U$ is dense in $U.$" I wonder what it ***is***. – 2017-01-31
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    The definition of "$E$ is nowhere dense" is that $\text{int}(\overline{E}) = \emptyset$ – 2017-01-31

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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.