# Proving a sequence has a subsequence with limits

Having trouble figuring out the following problem/proof.

If we let $(a_n)$ be a bounded sequence. I am trying to prove that $(a_n)$ has a subsequence $(a_{n_k})$ with $$\lim_{k\to\infty}a_{n_k}=\limsup a_n$$

I know that the following about a limsup,

Let $(s_n)$ be a sequence in $R$. We define $$\limsup\ s_n = \lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$

• @Michael Hardy thank you for the edit – user123 Feb 10 '17 at 0:13
• @Michael Hardy would you be able to help me? – user123 Feb 10 '17 at 0:22
• @MichaelHardy yes I typed that in wrong, I have edited my question – user123 Feb 10 '17 at 0:30

Let $s_{N}=\sup\{a_{n}:n>N\}.$ Then we know that for every $\varepsilon>0,$ and for every $N>0,$ there is some $n>N$ such that $s_{N}-\varepsilon<a_{n}\leq s_{N},$ by definition of the sup. For $\varepsilon=1/k,$ let $n_{k}$ be some index $>\max\{k,n_{k-1}\}$ such that this property holds, i.e., $$s_{k}-1/k<a_{n_{k}}\leq s_{k}\text{ for all }k\geq 1.$$ Then taking limits as $k\rightarrow\infty,$ we see that $$\lim\sup_{n} a_{n}\leq \lim\inf_{k} a_{n_{k}}\leq \lim\sup_{k} a_{n_{k}}\leq \lim\sup_{n} a_{n},$$ recalling that $\lim_{k\rightarrow\infty}s_{k}=\lim\sup_{n}a_{n}.$ Then $\lim_{k\rightarrow\infty}a_{n_{k}}$ exists and equals $\lim\sup_{n}a_{n},$ since $\lim\inf_{k}a_{n_{k}}=\lim\sup_{k}a_{n_{k}}=\lim\sup_{n}a_{n}.$ This completes the proof.
• are those n k k n letters that are floating supposed to be like $limsup_n$? – user123 Feb 10 '17 at 1:07
• so this proves that I that $(a_n)$ has a subsequence $(a_{n_k}) with $$\lim_{k\to\infty}a_{n_k}=\limsup a_n$$? – user123 Feb 10 '17 at 1:14 • Yes, the subsequence identified above has this property, since the second set of displayed inequalities force$\lim\inf_{k}a_{n_{k}}=\lim\sup_{k}a_{n_{k}}=\lim\sup_{n}a_{n},$which means the limit of$a_{n_{k}}$exists and equals$\lim\sup_{n}a_{n}.\$ – RideTheWavelet Feb 10 '17 at 1:22