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We have a compact set $X$ and a sequence $\{x_n\} \subset X$, clear enough all accumulation points of the sequence are in X. If the sequence has only one accumulation point, is it necessarily its limit, and if yes, then how to prove it?

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    $\begingroup$ Are you familiar with $\liminf$ and $\limsup$? $\endgroup$ Mar 21, 2015 at 17:31
  • $\begingroup$ @TimRaczkowski Yes I am, thank I see now. $\endgroup$
    – zesy
    Mar 21, 2015 at 17:32

2 Answers 2

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It depends on how you define accumulation points for a sequence.

If you consider the accumulation points of the set of values of the sequence, then the assertion is false. Consider the sequence in the compact set $[0,1]$ defined by $$ x_n=\begin{cases} 1/n & \text{if $n$ is odd,}\\ 1 & \text{if $n$ is even.} \end{cases} $$ Then the only accumulation point of the set $\{x_n:n\in\mathbb{N}\}$ is $0$, but the sequence doesn't converge.

If you define an accumulation point of a sequence as a point $p$ such that the sequence visits infinitely often every neighborhood of $p$, then the assertion is true. With just one accumulation point $p$ in this sense, you get that $\liminf_{n\to\infty}x_n=p=\limsup_{n\to\infty}x_n$.

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  • $\begingroup$ What view on the accumulation point is more widespread in math? $\endgroup$
    – zesy
    Mar 22, 2015 at 10:22
  • $\begingroup$ @SergeyZykov Personally I have used accumulation point only for sets. IIRC, my textbook at university used limit point for the second meaning. Terminology is not fully agreed upon, unfortunately. $\endgroup$
    – egreg
    Mar 22, 2015 at 10:28
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There is a theorem saying that $\liminf x_n = \inf\{z_n\}$ and $\limsup x_n = \sup\{z_n\}$ where $\{z_n\}$ is the set of accumulation points of $\{x_n\}$.

In our case $\{z_n\}$ has only one point, it means that $\liminf x_n = \limsup x_n$ and it means (by yet another theorem) that $\lim x_n$ exists and $\lim x_n = \liminf x_n = \limsup x_n$.

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