# Show that $\{x\in X\mid \forall$ open set $U, x\in U \implies A\cap U\neq\emptyset\} \subseteq A\cup A'$.

$A'$ is the derived set, i.e. $A'=\{x\in X\mid x\in\overline{A\setminus \{x\}}\}$.

I'm stuck on this because I literally ran out of ideas.

Let $x\in S:= \{x\in X\mid \forall$ open set $U, x\in U \implies A\cap U\neq\emptyset\}$. Then for each open set $U$ such that $x\in U$, $A\cap U \neq\emptyset$.

Suppose for a contradiction, $x\notin A\cup A'$, then $x\notin A$ and $x\notin A'$. But then so what?

Let $x\in X$ such that $x\in U$ implies $A\cap U\ne\emptyset$, for every open subset $U$ of $X$. To show $x\in A\cup A'$, we assume $x\not\in A$ and show $x\in A'$. Since $x\not\in A$, we know $$x\in A' \iff x\in\overline{A\setminus\{x\}}=\overline{A}.$$ Thus, in order to conclude $x\in\overline{A}$, we just need to prove that every closed set containing $A$ also contains $x$. To this end, suppose by contradiction that $F$ is a closed set such that $A\subseteq F$ and $x\not\in F$. Then there is an open set $U$ such that $x\in U \subseteq F^c$. But this implies $U\cap A\subseteq U\cap F=\emptyset$, which contradicts our assumption on $x$. Therefore $$x\in \bigcap\{F\subseteq X\ \text{closed} \mid A\subseteq F\} = \overline{A}$$ and consequently $x\in A'$ as desired.
• I'm not sure if $\overline{A\setminus\{x\}}$ is a good definition for $A'$, because this set notation seems to suggest that $A'$ is different for different $x$. – Sid Caroline Aug 4 '17 at 5:43
• But why is it that $x\in \overline{A\setminus\{x\}} \iff x\in \overline{A}$? – Sid Caroline Aug 4 '17 at 6:02
• Because we are assuming that $x\not\in A$, so that $A\setminus\{x\}=A$. – John Griffin Aug 4 '17 at 6:03