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Let $D$ be a dense in $X$. Prove that for every open set $U\subseteq X$, $$\newcommand{\cl}{\operatorname{cl}}\cl (D \cap U) = \cl(U)$$

For my solution, what I did is by showing that the $\cl(D \cap U)$ is contained in $\cl(U)$ and vice versa.

I've done the $\subseteq$. I have trouble in the $\supseteq$ part.

or is their an easier solution where I don't need the inclusions?

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I have added a \cl command to be available on this page. – Asaf Karagila Nov 19 '12 at 16:15

Fix $x$ an element of the closure of $U$, and $V$ a neighborhood of $x$. We have to show that $D\cap U\cap V$ is not empty. Let $y\in V\cap U$ (why does such a $y$ exists?). As $D$ is dense in $X$ and $V\cap U$ is a neighborhood of $y$, using the definition of the closure...

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Because $D\cap U\subseteq U$ we immediately have that $\cl(D\cap U)\subseteq\cl(U)$.

The other inclusion is that if $x\in\cl(U)$ then every open environment $V$ of $x$ is such that $V\cap U\neq\varnothing$. However $D$ is dense therefore $D\cap V\neq\varnothing$ as well, so every open environment of $x$ meets $D\cap U$ as well. Therefore $x\in\cl(D\cap U)$.

Therefore we have $\cl(D\cap U)=\cl(U)$.

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"there is (?) an open environment $V$" – Stefan Hamcke Nov 19 '12 at 16:19
@Stefan: Err, yeah. I was out of focus for a moment there. :-) Thanks! – Asaf Karagila Nov 19 '12 at 16:21

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