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Is this the correct thinking process?

I am thinking about a proposition that says a set is closed if it contains all of its limit points: I am just looking at --> (this direction)

I suppose that the closed subset is called $X$ of metric space $M$ and that an arbitrary limit point $a$ is not in the set. Since $X$ is closed then the complement is open. Since the complement is open, there exists an open ball of any point in the complement. If we look at the limit point in the complement we see that it does not contain an infinite number of points in $X$ since the points it contains are in the complement of $X$. So, then $a$, the limit point, has to be in $X$. So a closed set contains all of its limit points?

Am I thinking on the right tract? Thanks!

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  • $\begingroup$ What kind of a topological space do you have in mind? The notion of open ball only makes sense for metric spaces (and maybe some related kinds of spaces). $\endgroup$ – Travis Willse Apr 21 '15 at 4:21
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    $\begingroup$ A metric space, sorry forgot to add that in. X is a subset of a metric space M $\endgroup$ – michaelbaes Apr 21 '15 at 4:21
  • $\begingroup$ And what definition of "closed" are you using? Because your proposition is the definition of closed in some books. $\endgroup$ – Thomas Andrews Apr 21 '15 at 4:25
  • $\begingroup$ Closed if the complement is open $\endgroup$ – michaelbaes Apr 21 '15 at 4:26
  • $\begingroup$ Maybe I'm just having trouble parsing this so early in the morning over so little sleep, but "a set is closed if it contains all of its limit points" means that if a set contains all its limit points, then the set is closed. And you show the other direction, which makes no sense unless this is an "if and only if" proposition. $\endgroup$ – Asaf Karagila Apr 21 '15 at 4:30
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You are right. The proof can be simple as this:

Suppose that $x\notin X$ and $x$ is the limit point of $X$. Pick $U=X^{c}$ which is obvious contains $x$ and we have $U\cap X=\emptyset$. A contradiction!

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  • $\begingroup$ Awesome thanks!! $\endgroup$ – michaelbaes Apr 21 '15 at 4:27
  • $\begingroup$ why does the intersection being null make it a contradiction. I understand why it is null, but not why it tells us the limit point $x$ has to be in $X$ $\endgroup$ – michaelbaes Apr 21 '15 at 4:31
  • $\begingroup$ What is the meaning of "a limit point"? $\endgroup$ – Paul Apr 21 '15 at 4:33
  • $\begingroup$ Any open ball with center $x$ contains an infinite number of points in the set $X$ $\endgroup$ – michaelbaes Apr 21 '15 at 4:34
  • $\begingroup$ Should if be $U$ n $x$ = null, with a lower case x, not a capital $X$ $\endgroup$ – michaelbaes Apr 21 '15 at 4:35

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