Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show $A\subset (X,d)$ nowhere dense $\iff$ $(\overline{A})^o=\emptyset$.

My attempt: $A$ nowhere dense $\implies$ $(\overline{A})^c$ is dense in $X$. Then, $(A^c)^o$ is dense in $X$ (previously proven equivalence). Then, $\overline{((A)^c)^o}=X\implies\overline{((A)^c)^o}^c=\emptyset$. Then I have to show $\overline{((A)^c)^o}^c=(\overline{A})^o$. Am I going about this the right way?

share|cite|improve this question
Maybe you should include your definition of nowhere dense. – Sam Mar 4 '12 at 23:19
Here is convenient to specify what definition of nowheredense are you using. The claim in the title is often used as the definition of nowhere dense set. – leo Mar 4 '12 at 23:21
$A$ nowhere dense $\implies \bar{A}^c$ dense in $X$. – Emir Mar 5 '12 at 1:45
up vote 1 down vote accepted

Suppose that $\overline{A}$ contains an open set $U$. But because $A$ is nowhere dense there is open $V\subseteq U$ so that $V\cap A =\emptyset$. Hence no point of $V$ is a limit point of $A$, and hence no point of $V$ is in the closure of $A$. This is a contradiction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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