Let $(E, \tau)$ be a topological space,
For $U,V \in \tau $ st $U \cap V = \emptyset$. Prove $int(cl(U)) \cap int(cl(V)) =\emptyset$
I know $cl(A) = \displaystyle \bigcap_{A \subset S}S$ so $cl(A) \cap cl(B) = \displaystyle \bigg(\bigcap_{A \subset S}S\bigg) \cap \bigg(\displaystyle \bigcap_{B \subset T}T\bigg) $ where $S,T \subset E$ are closed
now need $int\displaystyle \bigg(\bigcap_{A \subset S}S\bigg) \cap int\bigg(\displaystyle \bigcap_{B \subset T}T\bigg)$
And $\displaystyle int(C) = \bigcup_{O \subset C}O, O$ is an open subset of $C \subset E$
So we get $\displaystyle \Bigg(\bigcup_{O \subset \bigcap_{A \subset S}S}O\Bigg) \cap \Bigg(\bigcup_{P \subset \bigcap_{B \subset T}P} P\Bigg)$ , $O \subset cl(A), P \subset cl(B)$ : open subsets
Also, for two sets the interior of the intersection equals the intersection of the the interiors. so
$\displaystyle \Bigg(\bigcup_{O \subset \bigcap_{A \subset S}S}O\Bigg) \cap \Bigg(\bigcup_{P \subset \bigcap_{B \subset T}P} P\Bigg) = \bigcup_{U \subset (\bigcap_{A \subset S}S)\, \cap \,( \bigcap_{B \subset T}T)}U, U$:open
Not sure how to get this equal to $\emptyset$.