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Show that $\limsup \{a_n\}$ is the largest cluster point.

Definition: We call an extended real number a cluster point of a sequence $\{a_n\}$ if a subsequence converges to this extended real number.

I'm not sure how to 'unwrap' the definition of a cluster point to prove this.

This was my attempt. Let $l = \limsup \{a_n\}$. Suppose that there is an $x$ such that $x > l$ and it is a cluster point of $l$. If we let $\epsilon>0$, then $x-\epsilon > l$. There exists $n \geq N$ such that $x_n > x-\epsilon$. Let $\tilde{n}=\min\{n \geq N \text{ such that } x_n > x-\epsilon \}$.

I'm not really sure where to go from here.

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You are basically almost there. Your $x_n$ is a sequence of points that converges to cluster point $x$ right? This should scream contradiction as that means there are infinitely many $x_n$ with $x_n>l+\epsilon$ – Alex R. Sep 17 '12 at 5:35
up vote 5 down vote accepted

To see that $\limsup\{a_n\}$ is a cluster point, note that for each $n$ we can find some $a_{k_n}$ with $k_n\geq n$ such that $\sup\{a_n,a_{n+1},\ldots\}<a_{k_n}+1/n$. Thus $\lim\limits_{n\to \infty}a_{k_n}=\limsup\{a_n\}$.

To see that $\limsup\{a_n\}$ is the largest, suppose $x$ is a cluster point of $\{a_n\}$, so we have a subsequence $\{a_{k_n}\}$ such that $\lim\limits_{n\to\infty}a_{k_n}=x$. Thus for each $\epsilon>0$, we have some $N$ such that $n\geq N\implies a_{k_n}>x-\epsilon$. Hence $\limsup\{a_n\}\geq x-\epsilon$ for all $\epsilon>0$, so $\limsup\{a_n\}\geq x$.

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This is so succinct. It blows mine out of the water. – emka Sep 17 '12 at 5:51
A minor technicality: the sequence $\{k_n\}$ constructed in the first part is not guaranteed to be strictly increasing – David Jul 16 at 20:04

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