Prove: $A\cup A'$ is the smallest closed set containing $A$, i.e.if $F$ is closed and $A\subset F\subset A\cup A'$ then $F = A\cup A'$
Definitions: $A'$ is the set of all accumulation or limit points.
proof: Let $F$ be closed and $A\subset F\subset A\cup A'$. Let $a\in F$. Since $F$ is closed, then $a$ is an limit point for $F$. Thus, by the definition of limit point there exists an open set $G$ containing $a$ contains a different point $q$ of $F$, where $a\neq q$. So we have, $$ q\in G\subset F$$ but, $A\subset F$ and $a$ is an accumulation point of $F$, then $F = A\cup A'$
I am not sure if I am right, any suggestions would be greatly appreciated
Best Answer
It's not true that if $a\in F$ then it is a limit point.
You're starting from the wrong side: you have to show that points in $A\cup A'$ are also in $F$.
If $a\in A\cup A'$, then either $a\in A$ or $a\in A'$.
In the first case there's nothing to prove. In the second case, $a$ is a limit point of $A$, hence also a limit point of $F$. Since $F$ is closed, you have $a\in F$.
Note that if $A\subset B$, then all limit points of $A$ are necessarily limit points of $B$.