# How to prove that the limsup of a sequence is equal to its greatest subsequential limit?

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.

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.

• thanks, Ill think about how to do this – mary Jan 12 '15 at 0:44

$\limsup$ 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.

• thanks, do you think this way is a better way to do it or rewritten's answer? – mary Jan 12 '15 at 4:19
• I think either way is fine. – Tim Raczkowski Jan 12 '15 at 4:22
• 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". – rewritten Jan 12 '15 at 8:30

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.