5

I know this is an easy question, but I cannot demonstrate it properly.

Suppose by contradiction that $A \cap \bar{Y} \neq \emptyset$. Then $\exists \ x \in A \cap \bar{Y}$.

I need help formalizing this reasoning (or correcting if it is wrong)

By the definition of $A$ open, $\exists \ U \subset A$ open neighborhood of $x$. Then my idea is to prove that $U \cap Y \neq \emptyset$ because $U$ in somewhat sense "touches" $Y$, because it's a neighborhood of $x$. But i can't write down a formal demonstration. Any help is appreciated. Thanks^^

Martin Sleziak
  • 53,687
Riccardo
  • 7,401

2 Answers2

10

Suppose that $A$ is open, and $A\cap Y=\varnothing$. Let $F=X\setminus A$; then $F$ is closed, and $Y\subseteq F$. Now take closures: $\operatorname{cl}Y\subseteq\operatorname{cl}F=F$, since $F$ is closed. But this immediately implies that $\operatorname{cl}Y\cap A=\varnothing$.

Clive Newstead
  • 64,925
Brian M. Scott
  • 616,228
3

Another type of proof of it:

If $A \cap \overline {Y} \neq \emptyset$, then there exists $x \in A \cap \overline {Y}$. Since $x \in \overline {Y}$, which implies that for any open set $U$ containing $x$, $U \cap Y \neq \emptyset$, and hence $A \cap Y \neq \emptyset$. A contradiction!

Paul
  • 20,553