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

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.

• How are $X$, $x_n$, and $x$ related? – user251257 Jun 19 '20 at 23:16
• @user251257 I think it's safe to assume $x_n, x \in X$. Fiona -- What definition of compact is available to you? – Robert Shore Jun 19 '20 at 23:25
• Is your space a metric space? – Kavi Rama Murthy Jun 19 '20 at 23:25
• 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 $x$. – copper.hat Jun 19 '20 at 23:27
• @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. – Fiona Everdeen Jun 19 '20 at 23:31

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.

• His test is very interesting, but I cannot understand the construction of set $A$. – Fiona Everdeen Jun 20 '20 at 0:33
• @FionaEverdeen $A$ is the set of sequence points that are not in $O$, see the definition. What’s not to understand? – Henno Brandsma Jun 20 '20 at 6:34

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?
• 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. – Fiona Everdeen Jun 20 '20 at 0:16
• @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. – Brian M. Scott Jun 20 '20 at 0:18
• 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. – Fiona Everdeen Jun 20 '20 at 4:47
• @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. – Brian M. Scott Jun 20 '20 at 4:53

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.

• Sorry. Why $X-U$ is compact? – Fiona Everdeen Jun 20 '20 at 0:13
• 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 $X \setminus U$. – Robert Shore Jun 20 '20 at 1:36
• Excellent, but I can't find or demonstrate why if $X-U$ is compact then that intersection must have an accumulation point. – Fiona Everdeen Jun 20 '20 at 4:36
• math.stackexchange.com/questions/449764/… – Robert Shore Jun 20 '20 at 20:20