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$A$ is a subset of a topological space $X$, in which $U$ is open.

I'm asking because I was looking at these exercises (this is the last one), and it specifies that $X$ is Hausdorff.

Here's my attempt of a proof:

'$\Rightarrow$':

$\emptyset \subseteq (U \cap A) \subseteq (U \cap \overline{A})$.

'$\Leftarrow$':

For any point $x\in U \cap \overline{A}$, every open set $U_x\ni x$ is such that $U_x \cap A \neq \emptyset$. As $U\ni x$, $U\cap A\neq \emptyset$ follows.

I haven't used the fact that $X$ is Hausdorff though. Have I made any 'illegal' assumptions, or fallacious deductions?

man_in_green_shirt
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1 Answers1

6

The equivalence is true in any topological space. If $U\cap\overline A$ is nonempty then $\overline A\setminus U$ is a closed set (intersection of $\overline A$ and $U^c$) that is strictly smaller than $\overline A$, hence cannot contain all of $A$, hence $U$ must intersect $A$.

Hagen von Eitzen
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