13

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 $S$?

Tom
  • 1,079
  • 1
    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. – AlexR Sep 22 '13 at 06:55
  • 3
    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. – Zev Chonoles Sep 22 '13 at 06:57
  • @ZevChonoles so $S$ is not closed does not mean "S is open"? That is so confusing. – Tom Sep 22 '13 at 06:58
  • 1
    @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. – Zev Chonoles Sep 22 '13 at 07:00
  • @Tom: Ah, did you make this change to the excerpt? I was surprised that a book would make this mistake. – Zev Chonoles Sep 22 '13 at 07:03
  • @ZevChonoles The book did actually say "not closed" instead of "open" but I naively assumed them to be the same thing. But with that said, 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 $S$? – Tom Sep 22 '13 at 07:03
  • @Tom Consider $[1,2)$ in $\mathbb{R}$. It is not closed, but it is also not open as there is no ball around $1$ contained in $[1,2)$. –  Sep 22 '13 at 07:06
  • Sets are not doors. A set can be open, closed, open and closed, and neither open nor closed. This made Hitler angry ;-) – kahen Sep 22 '13 at 07:10
  • See $:$ http://en.wikipedia.org/wiki/Door_space . $;;;$ –  Sep 22 '13 at 07:10
  • @Tom I wrote an answer; do you know what a contraposition is? – AlexR Sep 22 '13 at 07:11
  • @kahen That was so funny! That is exactly how I felt. Thanks for sharing that. – Tom Sep 22 '13 at 07:19
  • @kahen: The Nazis did kill some topologists, most notably Hausdorff committed suicide to avoid the death camps,and Lindenbaum was killed in Vilnus (although the latter was more of a logician). – Asaf Karagila Sep 22 '13 at 07:25

4 Answers4

5

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.

AlexR
  • 24,905
  • Alex what do you mean by contraposition and did you mean the intersection $=$ instead of $\ne$? – Tom Sep 22 '13 at 07:14
  • Contraposition of "open" is "not open". It's some form of a negation. Also, my notation works. It basically means $$B_\epsilon(x) \not\subset A$$ – AlexR Sep 22 '13 at 07:19
  • Hmm, this may take me a few minutes to wrap my head around. – Tom Sep 22 '13 at 07:21
2

$S$ 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.

Kaa1el
  • 2,058
1

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 $\frac1n$.

Andrew Chin
  • 7,389
James Perry
  • 11
0

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.

Tolulope Rasheed
  • 1
  • 1