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Show that if $X$ is compact and $x$ is the only point of accumulation of the sequence $\{x_n\}$ then $x_n$ converges to $x$.

How could I prove it, I know that for the convergence to be fulfilled I have to prove it by the definition of convergence, but since I use the hypothesis of being the only limit point.

Pd: Disculpen la traducción, mi inglés no es muy fluido.

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  • $\begingroup$ How are $X$, $x_n$, and $x$ related? $\endgroup$ – user251257 Jun 19 '20 at 23:16
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    $\begingroup$ @user251257 I think it's safe to assume $x_n, x \in X$. Fiona -- What definition of compact is available to you? $\endgroup$ – Robert Shore Jun 19 '20 at 23:25
  • $\begingroup$ Is your space a metric space? $\endgroup$ – Kavi Rama Murthy Jun 19 '20 at 23:25
  • $\begingroup$ There is a useful technical lemma that says that $x_n \to x$ iff for any subsequences of $x_n$ there is further subsequence that converges to $x$. $\endgroup$ – copper.hat Jun 19 '20 at 23:27
  • $\begingroup$ @user251257 X is a topological space, $x$ is a limit point, and it is also the point to which the sequence $x_n$ must converge. $\endgroup$ – Fiona Everdeen Jun 19 '20 at 23:31
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Suppose $(x_n)$ does not converge to $x$, then there is some open set $O$ containing $x$, such that $X\setminus O$ is infinite, i.e. $M:=\{n: x_n \notin O\}$ is an infinite set. If $A:=\{x_n: n \in M\}$ is finite, then some $p \notin O$ occurs infinitely many times, and we have a new accumulation point of $(x_n)$, a contradiction. So $A$ is infinite and thus has an $\omega$-accumulation point in the compact subset $X\setminus O$ and again we have a new (not $x$) accumulation point of $(x_n)_n$. So we always get a contradiction and $x_n \to x$ after all.

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  • $\begingroup$ His test is very interesting, but I cannot understand the construction of set $A$. $\endgroup$ – Fiona Everdeen Jun 20 '20 at 0:33
  • $\begingroup$ @FionaEverdeen $A$ is the set of sequence points that are not in $O$, see the definition. What’s not to understand? $\endgroup$ – Henno Brandsma Jun 20 '20 at 6:34
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HINT: Show that if $U$ is any open nbhd of $x$, there are only finitely many $n\in\Bbb N$ such that $x_n\notin U$. To do this, suppose that there are infinitely many $n\in\Bbb N$ such that $x_n\notin U$. There are two possibilities:

  • There is some $y\in X\setminus U$ such that $\{n\in\Bbb N:x_n=y\}$ is infinite. Why is this impossible?
  • $\{x_n:x_n\notin U\}$ is infinite. Then $\{x_n:x_n\notin U\}$ is an infinite closed set with no accumulation point (why?); is this possible in a compact space?
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  • $\begingroup$ The first point why $ y $ would be another limit point other than $ x $ and that contradicts the fact of the uniqueness of the limit point. For the second point, there is what I'm thinking. $\endgroup$ – Fiona Everdeen Jun 20 '20 at 0:16
  • $\begingroup$ @FionaEverdeen: You’re right about the first point. For the second, it may help to notice that if that set has no accumulation point, then every $x_n$ in it has an open nbhd that contains no other point of the set. $\endgroup$ – Brian M. Scott Jun 20 '20 at 0:18
  • $\begingroup$ Thanks for the hint, but it didn't break any rules if I say that since we know that $X$ is compact and that implies that it is compact by limit point then $A = \{x_n: x_n \not\in U \}$ for being infinity has a limit point and we arrive at the same contradiction as in the first point. Also, now that I've analyzed it a lot, I can't understand why in the first point the set $ \{n \in \mathbb{N}: x_n = y \}$ exists. I imagine it has to do with the set $ A $ in that case it is finite, but I have not managed to see it. $\endgroup$ – Fiona Everdeen Jun 20 '20 at 4:47
  • $\begingroup$ @FionaEverdeen: The reason for the first case is that in principle we could have $A$ finite and $\{n:x_n\in A\}$ infinite. That is, we have to worry about the difference between a limit (or cluster) point of a sequence and a limit point of a set; for instance, the sequence $\langle 0,1,0,1,0,1,\ldots\rangle$ has two cluster points, even though the set $\{0,1\}$ has no limit points. Here that can’t actually happen, because that, as you said, would mean that at least one of the points of $A$ was another limit point of the sequence, but we have to check that. $\endgroup$ – Brian M. Scott Jun 20 '20 at 4:53
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If $x_n$ doesn't converge to $x$, then there's an open set $U$ such that $x \in U$ and infinitely many elements of $x_n$ fall outside of $U$. But $X \setminus U$ is compact, so that means $\{ x_n \} \cap (X \setminus U)$ has an accumulation point as well (If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$), contradicting the hypothesis that $x$ was the only accumulation point.

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  • $\begingroup$ Sorry. Why $X-U$ is compact? $\endgroup$ – Fiona Everdeen Jun 20 '20 at 0:13
  • $\begingroup$ Let $\{V_\alpha \}$ be any open cover of $X \setminus U$. Then $U \cup \{ V_\alpha \}$ is an open cover of $X$, so it has to have a finite subcover. Remove $U$ from that finite subcover of X and you have a finite subcover of $X \setminus U$. $\endgroup$ – Robert Shore Jun 20 '20 at 1:36
  • $\begingroup$ Excellent, but I can't find or demonstrate why if $ X-U $ is compact then that intersection must have an accumulation point. $\endgroup$ – Fiona Everdeen Jun 20 '20 at 4:36
  • $\begingroup$ math.stackexchange.com/questions/449764/… $\endgroup$ – Robert Shore Jun 20 '20 at 20:20

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