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I'm considering a divergent sequence $\{x_n\}$ in a compact set $K \subset R^n$.

Since it should have a convergent subsequence, it has one limit point at least. Then, since it is not convergent,

should it have another limit point? Can anybody show that clearly please?

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3 Answers 3

Let $\langle x_{n_k}:k\in\Bbb N\rangle$ be a convergent subsequence, say with limit $x$. Since the original sequence $\langle x_k:k\in\Bbb N\rangle$ does not converge to $x$, $x$ must have an open nbhd $U$ such that infinitely many terms of $\langle x_k:k\in\Bbb N\rangle$ are outside $U$. Thus, if $A=\{k\in\Bbb N:x_k\notin U\}$, then $A$ is infinite. Consider now the subsequence $\langle x_k:k\in A\rangle$: it’s a sequence in the compact set $K\setminus U$, so it has a convergent subsequence whose limit must also lie in $K\setminus U$ and therefore must be distinct from $x$.

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Since the sequence diverges, when you throw away the convergent subsequence (i.e. remove its elements from the sequence), you get another sequence. As it is defined on a compact set, this new sequence also has a limit point. Note that this new sequence need not be divergent, and so this process cannot necessarily be repeated indefinitely. For example, consider the sequence

$$1, -1, 1, -1, 1, -1, \cdots $$

in $[0,1]$.Throwing away the subsequence of $-1$'s gives us a convergent subsequence of $1$'s.

If the sequence converges, throwing away the convergent subsequence amounts to removing all the elements in the sequence, and so the result does not hold.

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As you say yourself, by the Bolzano-Weierstrass theorem, your sequence has at least one limit point.

Assume your sequence only has a single limit point $p$. Then every neighborhood of $p$ contains all but a finite number of the points of the original sequence. (Otherwise there would be another limit point.) Hence $p$ is actually the limit of the sequence.

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