Accumulation points, Topological spaces Show that if $(X,\tau)$ is a topological space and for each $x \in \overline{A}, {\left\lbrace x \right\rbrace}'$ is closed, then $A'$ is closed.
I know $\overline{A} = A \cup A'$. So $\overline{A} = \cup_{x \in A} {\left\lbrace x \right\rbrace} \bigcup \cup_{x \in A'} {\left\lbrace x \right\rbrace}$
Also $y \in {\left\lbrace x \right\rbrace}' \longleftrightarrow$ for each $E \in E_y, E \cap ({\left\lbrace x \right\rbrace}-{\left\lbrace y \right\rbrace}) \neq \emptyset$ 
And $X-{\left\lbrace x \right\rbrace}' \in \tau$, so every $E$ in which $y \in E$ and $X-{\left\lbrace x \right\rbrace}' \subseteq E$; then $E \in E_y$. Please help
 A: So suppose $\{x\}'$ is closed for all $x \in X$.
Let's try to prove $A'$ is closed. So let $x \notin A'$, and we need to find an open set $O_x$ that contains $x$, and is disjoint from $A'$.
First note that $x \notin A'$ implies that there is an open neighbourhood $U_x$ of $x$ such that $U_x \cap (A\setminus\{x\})  =\emptyset$, i.e. $U_x \cap A= \emptyset$ or $U_x \cap A = \{x\}$, or $U_x \cap a \subseteq \{x\}$.
Also, by definition $x \notin \{x\}'$, so $X\setminus \{x\}'$ is open and contains $x$. Now define $O_x = U_x \cap (X\setminus \{x\}')$, which is an open neighbourhood of $x$. 
Let $p \in O_x$. Suppose $p \in A'$. This would mean that 
$O_x$, as an open neighbourhood of $p$, intersects $A$ in a point $a \neq p$. As $a \in O_x \cap A \subseteq U_x \cap A \subseteq \{x\}$ we have $a=x$, but 
$p \in O_x \subseteq X\setminus\{x\}' $, hence $p \notin \{x\}'$ so there is an open neighbourhood $V_p$ of  $p$ that does not contain $x$ at all. But then $p \in V_p \cap O_x$ misses $A$ entirely, so $p \notin A'$. 
This shows, as required, that $O_x \cap A' = \emptyset$ and $A'$ is closed.
