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I have a very tricky problem that I'm having a hard time figuring out how to start. Basically, I want to prove that the supremum of the set of subsequential limits of a sequence is equal to the lim sup of the sequence.

So I have a sequence $S_n$. I want to show that its greatest subsequential limit (which could either be a real number, infinity, or negative infinity) is equal to the limit (as N goes to infinity) of the supremum of the set $X=\{S_n:n>N\}$, $$\lim_{N\to\infty} \sup \{S_n:n>N\}$$ which is the definition of limit superior. I'm having a hard time coming up with a way to go about this.

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You have to prove it in two steps.

First, prove that the greatest sequential limit cannot be greater than the limsup (that's easy, using reductio ab absurdum, suppose a sequential limit is greater than the limsup, and derive a contradiction).

Then, prove that the greatest sequential limit cannot be smaller than the limsup (explicitly build a sequence whose limit is greater than any real number strictly smaller than the limsup).

The only other option is that those two limits are equal.

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  • $\begingroup$ thanks, Ill think about how to do this $\endgroup$ – mary Jan 12 '15 at 0:44
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$\limsup$ of a sequence is uniquely characterized by two properties. Let $\{a_n\}$ be a sequence of real numbers, and $A=\limsup_{n\to\infty}a_n$.

For any $\epsilon>0$.

a)$\exists N$ such that $n\ge N\implies a_n<A+\epsilon$.

b)$\exists n$ such that $a_n>A-\epsilon$.

If $A^*$ is the set of subsequential limits, show that $\sup A^*$ satisfies these two properties.

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  • $\begingroup$ thanks, do you think this way is a better way to do it or rewritten's answer? $\endgroup$ – mary Jan 12 '15 at 4:19
  • $\begingroup$ I think either way is fine. $\endgroup$ – Tim Raczkowski Jan 12 '15 at 4:22
  • $\begingroup$ I think that both ways are essentially the same, @TimRaczkowski's hint is more formal, mine is more like "where to go to prove your claim". $\endgroup$ – rewritten Jan 12 '15 at 8:30
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Use the definition: $\lim \sup S_n=\lim_{n\to \infty}T_n$ where $T_n=\sup \{S_m:m\geq n\}.$ Assume $-\infty< L=\lim \sup S_n<+\infty.$

(I). Take any $M>L.$ There are only finitely many $n$ such that $S_n\geq (M+L)/2.$ Because otherwise $T_n\geq (M+L)/2$ for every $n,$ implying $\lim_{n\to \infty}T_n\geq (M+L)/2>L.$

So $(S_n)_n$ cannot have a subsequence converging to $M$ for any $M>L.$ Because otherwise there are infinitely many $n$ with $|S_n-M|<(M-L)/2,$ implying there are infinitely many $n$ such that $S_n\geq (M+L)/2.$

(II). Take any $N<L.$ There do exist infinitely many $n$ such that $S_n>N.$ Otherwise $T_n\leq N$ except for finitely many $n,$ implying $\lim_{n\to \infty}T_n\leq N<L.$

(III). By (II) with $N =L-2^{-1},$ take $n(1)$ such that $S_{n(1)}> L-2^{-1}.$ Recursively, by (II) with $N=L-2^{-j-1}$ take $n(j+1)>n(j)$ such that $S_{n(j+1)}> L-2^{-j-1}.$

The sequence $(S_{n(j)})_j$ converges to $L.$

Because if $\epsilon >0,$ then by (I) with $M=L+\epsilon$, take $K\in \mathbb N$ large enough that $n\geq K\implies S_n<L+\epsilon.$ And take $j_K\in \mathbb N$ large enough that $n(j_K)\geq K$ and $2^{-j_K}<\epsilon.$ Then $$j\geq j_K\implies |S_{n(j)}-L|<\epsilon.$$

Remark: $\lim \sup S_n$ is the the largest number $x$ such that, for every $r>0,$ the set $\{n: S_n\in [-r+x,r+x]\;\}$ is infinite.

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