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For my Topology final yesterday, I proved that the intersection of two open dense sets is again a dense set. I used the fact that a set $A$ being dense in a space $X$ is equivalent to $X - A$ having empty interior.

My proof, however, only seemed to use the fact that one of the two sets was open. My question is, can we get away with this? Given an open dense set $U_1$ and a dense set $U_2$, is $U_1 \cap U_2$ necessarily dense? The question is obviously not true if neither set is open, but I cannot think of a counterexample when only one set is open.

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Yes, it is sufficient that one of the dense sets is open.

Let $V$ be any nonempty open set. Then $V \cap U_1$ is a nonempty open set, since $U_1$ is dense and open, and since $U_2$ is dense, we have $(V\cap U_1) \cap U_2 \neq \varnothing$.

Thus, every nonempty open set has nonempty intersection with $U_1\cap U_2$, and that means precisely that $U_1 \cap U_2$ is dense.

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  • $\begingroup$ Thank you. I thought so, I just didn't have my proof in front of me and wasn't sure if I had implicitly used the fact that $U_2$ was open somewhere. $\endgroup$ – Thomas Credeur Dec 12 '13 at 0:29
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Suppose that $U$ is a dense open subset of $X$, and $D$ is any dense subset of $X$. Let $V$ be a non-empty open set in $X$; then $V\cap U$ is a non-empty open set, so $V\cap U\cap D\ne\varnothing$. Thus, $U\cap D$ is indeed dense in $X$.

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