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If $D$ is a dense subset of a topological space, $X$, and $O$ is an open subset of of $X$, then prove $\it{O} \subseteq \overline{D \cap O}$.

Here's a summary of what I've done/thought so far:

First, I said let $x \in \it{O}$. Now we want to show that $x \in \overline{D \cap O}$.

$D$ is given to be dense $\implies$ $\overline{D} = X \implies$ every neighborhood of every element of $X$ is a closure point of $D \implies$ every neighborhood of every element of $X$ intersects $D$.

Also, $O$ is given to open $\implies$ $\forall x \in X, \ \exists$ a neighborhood, $N_x$, such that $N_x \subseteq O$.

At this point, I was unsure of how to use these ideas to prove my result, so I looked at the possibility of proving this via a contradiction.

In other words, suppose $\it{O} \not\subseteq \overline{D \cap O} \implies \exists x \in O$ such that $x \not\in \overline{D \cap O} \implies x \in (\overline{D \cap O})^c$.

Now, I have proved a property that says $(\overline{A})^c = (A^{c})^o$. Using this result, I get $x \in [(D \cap O)^c]^o \implies x \in (D^c \cup O^c)^o$.

I have tried to play with the above result, but am unable to get any traction.

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Take $x$ in $O$. You want to show $x \in \overline{D \cap O}$, i.e., you want to show that every neighborhood of $x$ intersects $D \cap O$ non-trivially. Since $D$ is dense, you already know that every neighborhood of $x$ intersects $D$ non-trivially. Take a neighborhood $U$ of $x$, then $U \cap O$ is also a neighborhood of $x$, so $U \cap O \cap D \ne \varnothing$.

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Either approach will work. Nitrogen has shown how to make the direct approach work. For the indirect approach, suppose that $x\in O\setminus\operatorname{cl}(D\cap O)$. Let $U=O\setminus\operatorname{cl}(D\cap O)$; then $U$ is open. (Why?) $U\ne \varnothing$, since $x\in U$. Can $U$ contain any point of $D$? Such a point would have to be in $O$, since $U\subseteq O$, and not in $\operatorname{cl}(D\cap O)\supseteq D\cap O$; is that possible?

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$N_x\cap(D\cap O) = (N_x\cap D)\cap O$

We have: $N_x\cap D \neq \emptyset$ (as $D$ is dense) and $(N_x\cap D) \subset N_x \subset O$ (as $O$ is open) $\Rightarrow (N_x\cap D)\cap O$ is nonempty subset of $O$.

Thus, for every $x\in O$, $N_x\cap(D\cap O)\neq \emptyset \Rightarrow x\in \overline {D\cap O} \Rightarrow O \subset \overline {D\cap O}$

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