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I am trying to solve this problem, and i think i did something, but finally i couldn't get the conclusion. The question is:

  • Let $(X,d)$ be a metric space and let $A,B \subset X$. If $A$ is an open dense subset, and $B$ is a dense subset, then is $A \cap B$ dense in $X$?.

Well, i think this is true. We have to show that $\overline{A \cap B}=X$. Or in other words, $B(x,r) \cap (A \cap B) \neq \emptyset$ for any $x \in X$. Since $A$ is dense in $X$, so we have $B(x,r) \cap A \neq\emptyset$. That means there is a $y \in B(x,r) \cap A$. Which means there is a $y \in A$. And since $A$ is open we have $B(y,r_{1}) \subseteq A$ for some $r_{1} > 0$. Couldn't get any further. I did try some more from here on, but couldn't get it. Any idea for proving it or giving a counter example.

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Let $x$ be any point and $\varepsilon > 0$. Pick a point $a \in A$ with $d(a,x) < \varepsilon$. Since $A$ is open there is $\delta \leq \varepsilon$ such that $B_{\delta}(a) \subset A$. Now there is $b \in B_{\delta}(a) \cap B \subset A \cap B$ and $d(b,x) < 2 \varepsilon$. –  t.b. Jan 10 '11 at 10:28

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Good start. Now use that $B(x,r) \cap B(y,r_1)$ is an open set (and nonempty since it contains $y$) and therefore contains some point of $B$.

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Thanks. What i was trying was since $B(y,r_{1}) \subseteq A \Rightarrow B(y,r_{1}) \cap B \subset A \cap B$. And then i thought for a long while and failed to conclude. –  anonymous Jan 10 '11 at 11:54

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