# Proof that the closure of $S$ is the set of all limits of convergent sequences in $S$

I'd like to validate my proof of: the closure of $$S$$ in a metric space $$X$$ is the set of the limits of all convergent sequences in $$S$$. The definition of $$\bar{S}$$ that I am adopting is that it is the intersection of all the closed sets containing $$S$$. Let $$L$$ be the set of the limits of all convergent sequences in $$S$$.

We'd like to prove:

• $$L \subseteq \bar{S}$$, i.e., $$x \in L \Rightarrow x \in \bar{S}$$
• $$\bar{S} \subseteq L$$, i.e., $$x \in \bar{S} \Rightarrow x \in L$$

We prove both sides using contrapositives.

To prove $$L \subseteq \bar{S}$$, suppose $$x \notin \bar{S}$$. Then there exists an open ball $$B(x,r)$$ which is not in $$\bar{S}$$. Since $$\bar{S}$$ contains $$S$$, there is no $$x_n \in S$$ such that $$x_n \in B(x,r)$$. Thus, the $$r$$-neighbourhood of $$x$$ does not contain any $$x_n \in S$$, and is thus not a limit point of $$S$$. Thus, $$x \notin L$$, thus proving the contrapositive.

To prove that $$\bar{S} \subseteq L$$, suppose that $$x \notin L$$. Thus, $$x$$ is not a limit point of $$S$$, implying that there exists an open ball $$B(x,r_0)$$ such that it contains no $$x_n \in S$$. Thus $$B'(x,r)$$ (the complement of the open ball $$B(x,r)$$) contains $$S$$. Since $$B(x,r_0)$$ is open, $$B'(x,r_0)$$ is closed. Denote $$B'(x,r)=P_r$$. Similarly, all balls $$B(x,r):r do not contain $$x_n \in S$$. Thus all $$B'(x,r)=P_r$$ are closed and contain $$S$$. It follows that $$x \in P_r'$$.

Thus, we write: $$x \in \displaystyle\bigcup_{i \in (0,r)} P_i' \\ \Rightarrow x \in \left(\displaystyle\bigcap_{i \in (0,r)} P_i\right)' \\ \Rightarrow x \in (\bar{S})' \\ \Rightarrow x \notin \bar{S} \\$$

thus proving the contrapositive.

I'd like feedback on the correctness of this proof, please.

• Duplicate Commented Nov 19, 2021 at 9:33
• The duplicate proves one of the cases slightly differently (without using a contrapositive). I'd like to validate my method of proof, please. Commented Nov 19, 2021 at 9:52
• It's not clear how you find $B(x, r_0)$ in your argument. Commented Nov 19, 2021 at 9:58
• Since $x$ is not a limit point of $S$, there exists an open ball $B(x,r_0)$ such that it contains no $x_n \in S$ for some $r_0>0$. I have edited the question to clarify this. Commented Nov 19, 2021 at 10:45
• The $x_n$ do not "exist" yet, so how can you talk about them? Commented Nov 19, 2021 at 11:56

For the first contrapositive you proved, you've written a sentence

Thus, the $$r$$-neighbourhood of $$x$$ does not contain any $$x_n \in S$$, and and is thus not a limit point of $$S$$, thus, $$x \notin L$$

The part that I made bold is actually a restatement of what you had aimed to prove. But it's still correct. In order to prove rigorously that $$x \notin L$$ you must show that $$x$$ is not a limit point for any sequence in $$S$$, which is not done here. So I would say some argument like the following is needed to be added for the proof of the first part,

Let $$\{s_n\}$$ be a sequence in $$S$$. Since no point of $$S$$ is in $$B(x,r)$$, for all $$i \in \mathbb{N}$$ we have $$s_i \notin B(x,r)$$ and so $$d(x,s_i) > r$$. So $$x$$ is not the limit of $$\{s_n\}$$. Since we didn't use any property of $$\{s_n\}$$ we may say that $$x$$ is not the limit of any sequence in $$S$$. Therefore $$x \notin L$$.

For the second contrapositive, it is correct that there is some $$r_0$$ such that it contains no $$x_n \in S$$; But a proof is required if it's not a first course, stating that otherwise a sequence could have been constructed (how?) which would have converged to $$x$$, contradicting $$x \notin L$$.

From there on, also, your proof is correct; but I think using the $$P_i$$s has contributed to the complication of your proof. Note that actually $$\bigcup_{i \in (0,r)} P_i' = \bigcup_{i \in (0,r)} B(x,i) = B(x,r)$$ So the second implication, $$x \in \Big(\bigcap_{i \in (0,r)} P_i\Big)'$$, is that $$x \in B(x,r)$$. And you could have said that $$B(x,r)'$$ is a closed set containing $$S$$ but not $$x$$, therefore $$x \notin \overline{S}$$.

But overall your proof is fine.

The first part needs clarification: you want to show $$x \in L$$ implies $$x \in \overline{S}$$, so start by picking a sequence $$(x_n)$$ from $$S$$ so that $$x_n \to x$$ (witnessing $$x \in L$$) and then you start a proof by contradiction that $$x \in \overline{S}$$. (So now the $$x_n$$ don't come out of nothing later on). It is indeed true that $$x \notin \overline{S}$$ implies the existence of some $$r>0$$ so that $$B(x,r) \cap S = \emptyset$$ and this follows from the fact that $$X\setminus \overline{S}$$ is open and has $$x$$ as an (interior) point. You should maybe clarify that better, because now the claim stands without any justification (unless it's a lemma in your text or notes already?). Then the final part is a valid contradiction, now that the sequence already has been introduced.

As to the proof that $$\overline{S} \subseteq L$$, I think that's rather muddled in your write-up. You assume $$x \notin L$$ (but it's unclear how to use that negative fact) and want $$x \notin \overline{S}$$ somehow. I think it's invalid as written.

Why not go the more direct route? Take $$x \in \overline{S}$$. Note that for each $$n \in \Bbb N$$, $$B(x,\frac1n) \cap S \neq \emptyset$$ (that needs a little argument from the definition of $$\overline{S}$$ but is true), and pick $$x_n$$ in that intersection and show that in the end $$x_n \to x$$ witnessing directly that $$x \in L$$. Much better, IMO.

The basic lemma that you're in need of here (and often used as an alternative definition of the closure), for all $$x \in (X,d)$$ and $$S \subseteq X$$ we know:

$$x \in \overline{S} \iff \forall r>0: B(x,r) \cap S \neq \emptyset$$

• Regarding the clarification of $B(x,r) \cap S = \emptyset$, would it be enough to argue like this: since $X/ \bar{S}$ is open, $x$ has an open ball completely in $X/ \bar{S}$. Furthermore since $X/ \bar{S} \cap \bar{S}=\emptyset$, it follows that $B(x,r) \cap S = \emptyset$. Commented Nov 19, 2021 at 12:20
• @EttenMoor better formatted: as $X\setminus \overline{S}$ is open there is some $r>0$ so that $B(x,r) \subseteq X\setminus \overline{S}$ and as $S \subseteq \overline{S}$, a fortiori $B(x,r) \cap S = \emptyset$. Commented Nov 19, 2021 at 12:22
• Regarding the proof around $\bar{S} \subseteq L$, if $x \notin L$, it's not a limit point, is it not implied that there exists some $r_0>0$ for which there are no points $y \in S$ which are in the open ball $B(x,r_0)$? I am merely using the union of all such open balls with $r<r_0$ and saying $x$ exists in their union; implying that $x$ does not exist in their complement (which turns out to be the closure of $S$). Commented Nov 19, 2021 at 12:26
• Regarding the basic lemma you pointed out: $B(x,r) \cap S \neq \emptyset$; I'd appreciate if you can point me to a reference which lists (or proves) that lemma: I'm currently using Kreyszig's Functional Analysis, and do not see that implication. Commented Nov 19, 2021 at 12:27
• @EttenMoor It's in Munkres or any standard topology text book. Commented Nov 19, 2021 at 12:28