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.

  • $\begingroup$ 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

(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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.