Closure of a set in a metric space.

Question

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.

Answer 1

The idea in the first half of the proof showing $N_r(a)\subseteq A^c$ is correct. But in my opinion when showing that $N_r(a)$ is not contained in $(A')^c$ your reasoning lacks explanations. I would do it as follows.

Assume $N_r(a)\nsubseteq (A')^c.$ Then $N_r(a)\cap A'\neq\emptyset.$ So let $b\in N_r(a)\cap A'.$ Then there exists $t>0$ such that $N_t(a)\subseteq N_r(a)$ (why?). Therefore $N_t(b)\cap A\neq \emptyset$ (why?). So $N_r(a)\cap A\neq\emptyset,$ which is a contradiction. Hence $N_r(a)\subseteq (A')^c.$

Answer 2

So I assume that your definition of "$A$ is closed" is "$A$ contains all its limit points", or equivalently $A' \subseteq A$.

If we then define $\overline{A}=A \cup A'$ then indeed this set is closed:

$$\overline{A}' = (A \cup A')' = A' \cup A'' \subseteq A' \subseteq \overline{A}$$ using that in a $T_1$ space (thus certainly in a metric space) we always have $B'' \subseteq B'$ for all subsets $B$ and also $(C \cup D)' = C' \cup D'$ for all subsets $C,D$. Thus $B = \overline{A}$ obeys $B' \subseteq B$ so is closed.

Moreover, if $C$ is closed and $A \subseteq C $ then $A' \subseteq C' \subseteq C$, so $\overline{A}=A \cup A' \subseteq C$, so it follows that $\overline{A}$ is the smallest closed set that contains $A$ as a subset, another way the closure could have been defined.