Let $(X, \tau)$ be any topological space. Let ${O}$ be any open set, and let $V$ be any other set (not necessarily open or closed). Suppose that
$O \cap closure(V) \not= \emptyset$.
Is it true that
$O \cap V \not= \emptyset$
as well? Here is my proof attempt:
Let $x \in O \cap closure(V)$. Then $x \in closure(V)$ implies the existence of some net $(x_{\alpha})_{\alpha \in A}$ of elements of $V$ such that $x_{\alpha} \rightarrow x \in O$. Thus, for some $\hat{\alpha}$, we have $\beta \succeq \hat{\alpha}$ implies $x_{\beta} \in O$. Since such $x_{\beta}$ are elements of $V$, we have $x_{\beta} \in O \cap V \not= \emptyset$.
What do you think? Nobel prize? Thanks!
Here is the fact we need "if $x\in cl(A)$ then $U\cap A\neq \emptyset$ for every neighbourhood $U$ of $x$ " And the other side of the fact also true.
Take any $x\in O \cap cl(V)$. Then $x\in O$ and $x\in cl(V)$. By using the fact we get $O \cap V \not= \emptyset$