Prop: $G \subseteq X$ is open in $X$ $\implies$ $cl(G\bigcap cl(A)) = cl(G\bigcap A)$ for every $A \subseteq X$.

Question

Denote the closure operator of a topological space by $cl$.

Prop: $G \subseteq X$ is open in $X$ $\implies$ $cl(G\bigcap cl(A)) = cl(G\bigcap A)$ for every $A \subseteq X$.

Pf: Assume that $G$ is open in $X$ and $A \subseteq X$. It is trivial that $G \bigcap A \subseteq G \bigcap cl(A) \implies cl(G\bigcap A)\subseteq cl(G\bigcap cl(A))$. Further assume that $x \in cl(G\bigcap cl(A))$. Then for every $x \in U \subseteq X$, $U\bigcap (G\bigcap cl(A))\neq \emptyset$. Using the fact that $G$ is open, there exists $y \in U$ such that for $y \in V \subseteq X$, $V \subseteq G$. Since $(V\bigcap G)\bigcap cl(A) \neq \emptyset \implies V \bigcap cl(A)\neq \emptyset \implies V \bigcap A \neq \emptyset$, $U \bigcap (G\bigcap A)\neq \emptyset$ and $x \in cl(G \bigcap A)$ as required.

Does my proof look correct?

Answer 1

Prove the following highly useful lemma:
open U implies U $\cap$ cl K subset cl (U $\cap$ K).

Thus cl (U $\cap$ cl K ) subset
cl cl (U $\cap$ cl K) = cl (U $\cap$ K)
subset cl (U $\cap$ K).

Your proof fails because U and V are just any sets.