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 ?

• The argument: $K$ is open and not dense in $X$, thus $Int(Cl(K))= \emptyset$.... seems wrong.
– Surb
Commented Feb 11, 2015 at 7:21
• I like the notion of a dense sunset. Commented Feb 11, 2015 at 7:25

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 a nonempty 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.

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 ... ?

• Apparently the downvoter is a bit … dense. Commented Jan 12, 2022 at 3:33
• Brain M. Scott. Long time I did not see your answers or comments in MSE. I hope you are doing well.
– 00GB
Commented Jan 12, 2022 at 4:27
• @00GB: Thank you. I am, and I hope that you are, too. I’m not really active at MSE any longer, but I do try to respond to pings, and I occasionally find and correct typos (or worse errors!) in old answers. Commented Jan 12, 2022 at 8:54
• Brian M. Scott, we really missed you at MSE, specifically, in set theory and topology tags.
– 00GB
Commented Jan 12, 2022 at 16:38

recall that if $U$ is an open set of $X$ and $A$ a dense subset of $X$, then $\overline{U\bigcap A}=\overline{U}$.

Another approach:

It is easy to show that $$G\cap \overline{H}=G\subseteq \overline{G\cap H}$$. Therefore, by monotonicity of closure:

$$\overline{G}=X\subseteq \overline{\overline{G\cap H}}=\overline{G\cap H}.$$