$\def\Irr{\operatorname{Irr}}$ Let $\mathbb{R}$ be the space of real numbers. If $A$ is a dense set in $\mathbb{R}$ and $B \subset \mathbb{R}$. We know that $A\cap B$ need not be dense in $B$. By dense sets in $B$, I mean $\overline{A\cap B}=B$.

For example, take $A=\mathbb{Q}$ "the set of all rational numbers", and take $B=\left\{1\right\}\cup \Irr$, where $\Irr$ is the set of all irrational numbers. Then it is clear that $A$ is dense in $\mathbb{R}$; however, $A\cap B$ is not dense in $B$.

My question is that: Under what conditions, the statement will be true; i.e. when $A\cap B$ will be dense in $B$, if $A$ is a dense in $\mathbb{R}$?


The answer is positive if $B$ is open, as it has topology induced by the topology of $X$ (here $R$).

  • $\begingroup$ The more crucial question is: if a set $B$ satisfies the property that for all $D\subset \mathbb{R}$ dense, $B\cap D$ is dense in $B$, must $B$ be open? $\endgroup$ – guest Nov 4 '14 at 0:42
  • $\begingroup$ @guest At least for $\bf R$ not. A counterexample is $[0,1]$. $\endgroup$ – Przemysław Scherwentke Nov 4 '14 at 0:44
  • $\begingroup$ @Przemysław Scherwentke, If $B$ is open set, then how it will be equal to the closure of a set which is actually closed? $\endgroup$ – Sara Nov 4 '14 at 0:49
  • $\begingroup$ Closure in the subspace topology. $\endgroup$ – Matt Samuel Nov 4 '14 at 0:52
  • $\begingroup$ @user145405 As far as I understand the question, OP has in mind the relative topology. $\endgroup$ – Przemysław Scherwentke Nov 4 '14 at 0:52

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.