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?


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.