# A set is closed if and only if it contains all its limit points.

A set is closed if and only if it contains all its limit points.

Proof in book: Suppose $S$ is "not closed". We must show that $S$ does not contain all its limit points. Since $S$ is "not closed", $S^c$ is "not open". Therefore there is at least one element $x$ of $S^c$ such that every ball $B(x,\epsilon)$ contains at least one element of $S$ ...

Why is there at least one element $x\in S^c$ such that every open ball contains at least one element of the open set $S$?

• That's a border point. But without assumptions to $S$, this is false, since both $\emptyset$ and $\mathbb R^n$ are both open and closed. Commented Sep 22, 2013 at 6:55
• From the excerpt you've given, I'd say that proof is wrong. It seems like it wants to proceed by contradiction for the $\impliedby$ direction: it's supposing that a set $S$ is not closed, then it wants to show that $S$ does not contain all its limit points. The problem is that $$S\text{ is not closed}$$ is an entirely different thing than $$S\text{ is open}$$ Indeed, a set can be open and closed; a set can also be neither open nor closed. Commented Sep 22, 2013 at 6:57
• @ZevChonoles so $S$ is not closed does not mean "S is open"? That is so confusing.
– Tom
Commented Sep 22, 2013 at 6:58
• @Tom: It's an unfortunate choice of terms, but it is so established that there's no chance of it changing. Best to get used to it. See here on Wikipedia for instance. Commented Sep 22, 2013 at 7:00
• @Tom: Ah, did you make this change to the excerpt? I was surprised that a book would make this mistake. Commented Sep 22, 2013 at 7:03

## 4 Answers

After the edit, everything is fine. Since $$A \text{ open} \Leftrightarrow A^C \text{ closed}$$ $S^C$ is not open, but an open set is characterised by $$\forall\ x \in A\ \exists\ \epsilon > 0 \ : \ B_\epsilon (x) \subset A$$ The contraposition is $$\exists\ x \in A\ : \forall\ \epsilon > 0 : B_\epsilon(x) \cap A^C \neq \emptyset$$ Which is what is stated in the book.

• Alex what do you mean by contraposition and did you mean the intersection $=$ instead of $\ne$?
– Tom
Commented Sep 22, 2013 at 7:14
• Contraposition of "open" is "not open". It's some form of a negation. Also, my notation works. It basically means $$B_\epsilon(x) \not\subset A$$ Commented Sep 22, 2013 at 7:19
• Hmm, this may take me a few minutes to wrap my head around.
– Tom
Commented Sep 22, 2013 at 7:21

$S$ is closed iff $S=\overline{S}$. There is a theorem says $x\in\overline{S}\leftrightarrow\forall U(U\mathrm{\ open}\wedge x\in U\rightarrow U\cap S\neq\emptyset)$. We show $\overline{S}=S\cup S'$ ($S'$ denote the derived set). For $\supseteq"$, note $S\subseteq\overline{S}$ and if $x\in S'$, then every neighborhood of $x$ intersect $S$ in a point different from $x$, so using the previous theorem, $x\in\overline{S}$. For $"\subseteq"$, suppose $x\in\overline{S}\setminus S$, by the previous theorem, every neighborhood of $x$ intersects $S$, but $x\notin S$ implies it must intersect $S$ in a point different from $x$, so $x\in S'$.

So $S$ closed implies $S'\subseteq S$. and if $S'\subseteq S$, we infer $S=\overline{S}$. QED

Reference: Theorem 17.6, 17.7 of Topology by Munkres, James.

Suppose there exists a sequence $$x_n$$ in $$S$$ converging to an element $$x$$ not contained in $$S$$. Then as a consequence of the definition of convergence, every ball centered at $$x$$ contains infinitely many elements of $$x_n$$ (we only need at least 1 for the following argument). Since $$x_n$$ are not elements of the complement of $$S$$, this means that the complement of $$S$$ is not open. This means that $$S$$ is not closed.

Conversely: Suppose $$S$$ is not closed. This means the complement of $$S$$ is not open. This means that there exists an $$x$$ contained in the complement of $$S$$ such that there doesn't exist a ball centered at $$x$$ that is contained in the complement of $$S$$. Hence every ball centered at $$x$$ intersects with $$S$$. Hence there exists a sequence $$x_n$$ that converges to $$x$$ defined by taking an element from $$S$$ contained in the ball centered at $$x$$ with radius $$\frac1n$$.

For the converse. Assume that S contains its limit points.so if x is in S' then x is in S. Recall that Closure(S)= S U S'. Since Closure of S is the smallest closed set containing S i.e S ⊆ Closure(S). Thus, S is closed.

• Please use $\rm \LaTeX$. Commented Feb 19, 2017 at 8:50