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Prove that if every convergent sequence in a set $C \subset \mathbb{R}^n$ converges to a point in $C$, then $C$ is closed.

The problem with this is that the definition my book uses is not the one with the limit points, it is this one.

A set is closed if and only if it's complement is open.

I don't know how to show it with that definition.

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2 Answers 2

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Proof sketch:

To show that $C$ is closed, you need to show $C^c$ is open. As usual that means you fix an $x \in C^c$ and you need to show that for some $\epsilon > 0$, the ball of radius $\epsilon$ around $x$, $B(x, \epsilon)$, is contained in $C^c$.

Suppose this were not the case. Then $B(x, 1/n)$ would always contain a point, call it $x_n$, in $C$. Then show $x_n$ is a sequence in $C$ but it converges to a point outside of $C$, which is a contradiction.

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  • $\begingroup$ I think I can do this. If I consider the candidates for the limits of any convergent sequence, the limit points of $C$ are always candidates. If, by assumption, that EVERY convergent sequence converges to a point in $C$, so the limit points must be in $C$ and i can consider $C^c$ and find such a ball that contains one of these points. $\endgroup$ Feb 23, 2014 at 4:27
  • $\begingroup$ can you explain how you produced this $B(x,1/n)$ ? where did the n come from? $\endgroup$ Feb 23, 2014 at 5:12
  • $\begingroup$ You're trying to show that for some $\epsilon > 0$, $B(x, \epsilon)$ is contained in $C^c$. So you assume towards contradiction that for EVERY epsilon, it is NOT contained in $C^c$. In particular, for $\epsilon = 1, 1/2, 1/3, 1/4, \ldots$, that means that $B(x,1)$ contains a point in $C$, $B(x,1/2)$ contains a point in $C$, and so on. So $B(x, 1/n)$ contains a point in $C$; call this point $x_n$. Then show that $x_n \to x$, but that is a contradiction because $x$ is not in $C$, and every sequence in $C$ has to converge to a point in $C$. $\endgroup$ Feb 23, 2014 at 5:37
  • $\begingroup$ thank you, that helps a lot. $\endgroup$ Feb 23, 2014 at 5:50
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Lemma: A set contains all of its limit points if and only if the set is closed.

Proof: $\Longleftarrow$ Let $C$ be a closed set. By definition, $C^c$ is open. Let $x$ be a limit point of the set $C$. If no such point exists, $C$ contains all of its limit points and hence is closed. Assuming $x$ exists, either $x\in C$ or $x\in C^c$. If $x \in C^c$, because $C^c$ is open, then there exists a $r>0$ such that $B_r(x)\subseteq C^c$. However, this implies that $B_r(x)\cap C=\phi$. However, this violates the fact that $x$ is a limit point of $C$. Thus, $x \not\in C^c$. Thus, $x\in C$. Hence, every limit point of $C$ is contained in $C$. Hence, $C$ is closed.

$\Longrightarrow$ Now, let $x$ be a limit point of $C$. We know that $x\in C$. Now, let $y \in C^c$. Because, $y \not \in C$, $y$ is also not a limit point of $C$, thus, there exists a $r>0$ such that $B_r(y)\cap C=\phi$ or $B_r(y)\subseteq C^c$. Thus, $C^c$ is open. Thus, $C$ is closed.


Proof (By Contraposition): Let $C$ not be closed. Thus, there is an $x$ which is a limit point of $C$ such that $x\not \in C$. Now, because $x$ is a limit point of $C$ for every $n\in \mathbb{Z}$, $B_{1\over n}(x)\cap C$ is non-empty. Let $\beta_n$ be a an element of this intersection for a given $n$. Define a sequence with terms $\beta_n$. This sequence is convergent and converges to $x$ (By Design). However, $(x_n)$ does not converge in $C$. $$\blacksquare$$

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  • $\begingroup$ Is this example saying that I need the definition of limit points of a closed set to complete the proof? $\endgroup$ Feb 23, 2014 at 4:08

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