A Fundamental Property of Metric Spaces ...

Question

Let $(X,d)$ be a metric space and $A\subset X$ and also suppose that $G$ is open in $X$ prove the identity:
$$ \overline {G\cap A}=\overline {G\cap \overline A} $$

Proposition: The intersection of a dense set and a open dense set is a dense set.

Answer 1

We actually don't need any metric, just a topology.

The inclusion from left to right is obvious, as $A \subset \overline{A}$, so we take the closure of a possibly larger set, which is larger.

So suppose $x \in \overline{G \cap \overline{A}}$, we need to show it is in $\overline{G \cap A}$. To this end, take any open neighbourhood $U$ of $x$, we will show it intersects $G \cap A$. We do know that $U$ intersects $G \cap \overline{A}$, as $x$ is in its closure. So pick $y \in U \cap (G \cap \overline{A})$. Then $y$ has $U \cap G$ as an open neighbourhood (here we use that both $U$ and $G$ are open!), so $U \cap G$ intersects $A$, as $y \in \overline{A}$. But this just says that indeed $U \cap (G \cap A) = (U \cap G) \cap A$ is non-empty, which was what we needed to show.

The final proposition follows, as we can take $A$ to be a dense set, so $\overline{A} = X$ and then we have $\overline{G \cap A} = \overline{G \cap X} = \overline{G}$. This says that $G \cap A$ is dense in $G$. If moreover $G$ is dense, we see that the closure equals $X$, so $G \cap A$ is dense when $A$ is dense and $G$ is both open and dense.

Answer 2

Let $x\in X$ and $U$ be open neighborhood of $x$. Assume $x\in\overline{G\cap \overline{A}}$. There exists $y\in G\cap \overline{A}$ such that $y\in U$. Since $U$ and $G$ are open, $y$ has neighborhood $V\subset U\cap G$. As $y\in\overline{A}$, there exists $z\in A$ such that $z\in V$. We have $V\subset G$, so $z\in A\cap G$. Also, $V\subset U$, so $z\in U$. Hence, $x\in\overline{G\cap A}$.