# Closure of a set in a metric space.

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.

• You start by basically assuming a not in X and proceed to make contradicition; thusly proclaiming X is the whole space, which usually it isn't. For example when A is empty. Jan 13, 2019 at 3:18
• You dont need the contradiction. Once you show that $N_r(a)$ exists and misses A, you've shown that X$^c$ is open. Thus X is closed. Jan 13, 2019 at 3:19
• @WilliamElliot but if A is empty then X is closed. Jan 13, 2019 at 3:21
• @WilliamElliot what do you suggest I do differently? Jan 13, 2019 at 3:22
• Topologically limit points are nearly useless. It is weird that limit points play such a central role in metric space theory. They are clumbsy and their use is not helpful as an introduction to topology. A better alternative for analysis than limit points is punctured nhoods, to be used only as needed for those aspects of analysis that topology is less concerned about. @mathsssislife Jan 13, 2019 at 9:11

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 $$N_r(a)\subseteq (A')^c.$$

• Would this be better: Let p$\in$ $(E \cup E′)^c$ which implies that p $\notin$ E and p $\notin$ E'. Since p$\notin$E', $\exists$ $N_r(p)$ : $\forall$ q$\in$ Nr(p) , q $=$ p or q$\notin$ E which is equivalent to Nr(p) $\cap$ E $=$ $\emptyset$ , therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $\subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $\subset$ E' $\implies$ p $\in$ E' which is a contradiction, therefore $N_r(p)$ $\subset$ $E^c$ $\cup$ E' which means that it is open, therefore the closure is closed. Jan 13, 2019 at 23:46

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.