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$p$ is a cluster point of $S\subset M$ if each neighborhood of $p$ contains infinitely many points. Here is my confusion, a cluster point is also a limit point of $S$, right?

If so, then how does the sequence $((-1)^n)$, ${n\in \mathbb N}$ has two cluster points namely $1, -1$ especially since the sequence does not have a limit as n approaches infinity.

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    The phrase "limit of a sequence" cares about the order of the terms in the sequence, but the phrase "cluster point" doesn't. Your sequence doesn't have a limit because it oscillates for arbitrarily large $n$. But it does have cluster points, as you say. It may help you to prove the following: if $a$ is a cluster point of $\{a_n\}$, then there is some subsequence $\{a_{n_k}\}$ with $\{a_{n_k}\} \to a$.2012-11-01
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    I mean if $a$ is a cluster point of $\{a_n\}$, then $a$ is a limit point of the sequence; thus the subsequence must converge to a. @countinghaus: If you can, just expand on the topic. As yo u can see, I am pretty lost.2012-11-01
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    @user43901: "[limit point](http://en.wikipedia.org/wiki/Limit_point)" and "[limit](http://en.wikipedia.org/wiki/Limit_%28mathematics%29)" are two different things.2012-11-01
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    @DejanGovc: elaborate on that please. Isn't a limit point $p \text{ of } S$ the limit of a sequence say $(p_n)$ that converges in $S$? I just don't see the connection; hence, if anyone can take their valuable time and explain this scenario more fulling instead of one-liners, it will be much beneficial for the understanding.2012-11-01
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    @Dejan’s comment isn’t really a one-liner: he’s directing you to the definitions of the two comments and suggesting that you see for yourself that they really aren’t the same thing.2012-11-01

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