I have a question with regards to the proof of the following: Let $X = A \cup A'$ be the closure of a set. $A'$ is the set of limit points of $A$. I wish to show that the closure is closed. I would very much like your opinions on the proof.
Let $a\in A^c\cap (A')^c$. Since $a$ is not a limit point of $A$ there exists a neighborhood of $a$ such that for every $q$ in the neighborhood of $a$, $q=a$ or $q\notin A$. This implies that $N_r(a)$ is contained within the complement of $A$. Now we would like to show that the neighborhood is contained within $(A')^c$, because then $N_r(a)$ $\subset$ $A^c$ $\cap$ $(A')^c$ which implies that the set is open, therefore the closure is closed. So assume that the neighborhood is not contained in $(A')^c$ this implies that $a$ $\in$ $A'$ because $N_r(a)$ $\subset$ $A'$, which is a contradiction.