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Suppose that a sequence $\{x_k\}$ has a clusterpoint (or more..) $c\in\mathbb{R}$. what conclusion, if any, can be drawn about either $\liminf x_k$ or $\limsup x_k$ ?

I don't know the conclusion... I think that when the sequence has only one cluster point then $\liminf x_k= \limsup x_k = c$. But if the sequence has many cluster points, I can't draw conclusion about either $\liminf x_k$ or $\limsup x_k$.

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2 Answers 2

There are several different (but equivalent) ways to define $\limsup$ and $\liminf$, so it all depends on which one you are familiar with.

One definition is that $\limsup x_k$ is the supremum of all points that are limits of subsequences of $x_k$. That is, $$\limsup x_k = \mathrm{sup}\Bigl\{ a\in\mathbb{R}\cup\{\pm\infty\}\Bigm| \text{there is a subsequence of }x_k\text{ that converges to }a\Bigr\}$$ (where we think of a subsequence as "converging to $\infty$" or "to $-\infty$" if the limits are equal to $\infty$ or $-\infty$). Similarly, one can define $\liminf x_k$ to be the infimum of all points that are limits of subsequences of $x_k$.

Note that a real number $a$ is a limit of a subsequence of $x_k$ if and only if $a$ is a cluster point for the sequence $x_k$. Given this, the fact that $c$ is a cluster point of $x_k$ tells you that $c$ belongs to that set for which $\limsup x_k$ is the supremum, and $\liminf x_k$ is the infimum. So what conclusion can you draw then?

If you have other definitions of $\limsup$ and $\liminf$, then please state the ones you know explicitly in your question.

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I understand that lim sup xk is supremum of the set of cluster points and lim inf xk is infumum ... but ... this is not conculsion ,,it is just definition ... so... can't draw conclusion... sorry –  Ji Yeon Park Mar 28 '11 at 16:21
    
@Ji Yeon Park: If you know this definition, and you know that $c$ is a cluster point, then what conclusion can you draw about the infimum of a set, if you know that $c$ is in the set? What conclusion can you draw about the supremum of a set, if you know that $c$ is in the set? –  Arturo Magidin Mar 28 '11 at 16:24
    
um... i think..the conclusion is lim infxk < c < lim sup xk... –  Ji Yeon Park Mar 28 '11 at 16:33
    
@Ji Yeon: Not quite; you have no warrant for the strict inequality. In general, if $a\in A$, then $\inf A\leq a\leq \sup A$; this is what you have here. And it follows rather from the definition of "infimum" and "supremum". What is that is giving you such great pause here? It's really supposed to be very easy, not a trick question at all. –  Arturo Magidin Mar 28 '11 at 16:36
    
oh I forgot "=" sorry.. and thank you very much for helping me sove this problem . thank you :) –  Ji Yeon Park Mar 28 '11 at 16:57

In general you have $\liminf x_k \leq c \leq \limsup x_k$. Note that $\liminf x_k$ may be strictly smaller than any of its clusterpoints ( $x_k = -k$ which even has no clusterpoints at all ) and a similar result obviously holds for $\limsup x_k$.

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