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 $$\limsup\ s_n = \lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$

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

  • $\begingroup$ are those n k k n letters that are floating supposed to be like $limsup_n$? $\endgroup$ – user123 Feb 10 '17 at 1:07
  • $\begingroup$ Yes. In inline equations, those appear as subscripts, but when they are displayed, they get tucked underneath, as with summation notation. $\endgroup$ – RideTheWavelet Feb 10 '17 at 1:12
  • $\begingroup$ so this proves that I that $(a_n)$ has a subsequence $(a_{n_k}) with $$$\lim_{k\to\infty}a_{n_k}=\limsup a_n$$? $\endgroup$ – user123 Feb 10 '17 at 1:14
  • $\begingroup$ 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 $\lim\sup_{n}a_{n}.$ $\endgroup$ – RideTheWavelet Feb 10 '17 at 1:22
  • $\begingroup$ where should I add that into the explanation? could you edit that in $\endgroup$ – user123 Feb 10 '17 at 1:25

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