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$\def\Irr{\operatorname{Irr}}$ 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 $\mathbb{R}$?

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

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  • $\begingroup$ 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

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