Intersection of two open dense sets is dense Let $X$ be a topological space and suppose that $H$ and $G$ are open dense subsets of $X$.Then show that $G \bigcap H$ is also an open dense subset of $X$.
My attempt :
Well since the finite intersection of open sets is open therefore, $G \bigcap H$ is also open.
I was trying to prove the other part by the method of contradiction :
Suppose on the contrary that $G \bigcap H$ is not dense. Then that implies that $Int(Cl(G \bigcap H)) = \phi$ .
Now, $Cl(G \bigcap H) \subset Cl(G) \bigcap Cl(H)$
so, $Int(Cl(G \bigcap H)) \subset Int(Cl(G) \bigcap Cl(H)) = Int(Cl(G)) \bigcap Int(Cl(H)) = X $.
How do i proceed further to get a contradiction ?
 A: Working with interiors and closures is messy; it’s easier to use the fact that a set $D$ is dense in $X$ if $D\cap U\ne\varnothing$ whenever $U$ is a non-empty open set in $X$. Thus, you want to show that if $U$ is a non-empty open set in $X$, then $U\cap(G\cap H)\ne\varnothing$. HINT: $U\cap G$ is a non-empty open set (why?), and $H$ is dense in $X$, so ... ?
A: You can't proceed further from that point since the inclusion you get at the end doesn't give you anything. I suggest a direct approach. To show that $G\cap H$ is dense in $X$, it suffices to show that $G\cap H$ intersects every nonempty open subset of $X$. To this end, let $V$ be an open subset of $X$. Since $G$ is open and dense in $X$, $G \cap V$ is a nonempty open set. Then since $H$ is dense in $X$, $H\cap G \cap V$ is nonempty, as desired.
A: recall that if $U$ is an open set of $X$ and $A$ a dense subset of $X$, then $\overline{U\bigcap A}=\overline{U}$.
A: Another approach:
It is easy to show that $G\cap \overline{H}=G\subseteq \overline{G\cap H}$. Therefore, by monotonicity of closure:
$$\overline{G}=X\subseteq \overline{\overline{G\cap H}}=\overline{G\cap H}.$$
