It is actually problem 4.13 in Folland's Real Analysis.
If $X$ is a topological space, $U$ is open in X, $A$ is dense in X, then $\overline{A\cap U} = \overline{U}$.
This looks obvious since intersection of an open set with a dense set is dense in the open set. But Folland's book doesn't include the concepts of dense in a subset. It only mentions dense in topological space $X$. So how to approach this problem?
Thanks!
Best Answer
One inclusion should be obvious: $A \cap U \subset U$, so that $\overline {A \cap U} \subset \overline U$.
For the other, let $x \in \overline U$.
1)By definition, we know that for every neighbourhood $W_x$ of $x$ intersects with $U$,
and so gives rise to a nonempty open set $W_x\cap U$.
2)By definition of dense, every nonempty open set intersects $A$.
Use these facts to show that every neighbourhood of $x$ contains a point of $A \cap U$, so that $x \in \overline{A \cap U}$.