...limit of a subsequence

[filter54321]

1. The problem statement, all variables and given/known data
Show that the lim sup of a bounded sequence is a limit of a subsequence.


2. Relevant equations
Sequence: Sn
Subsequence: Snk

3. The attempt at a solution
An existent lim sup means that at a large enough N, the subsequence could hug the bottom of the lim sup to within epsilon (e). I don't know how to formalize this notion.

The last two steps of the proof are likely
-e < Snk - lim sup Sn < e
|Snk - lim sup Sn| < e

[HallsofIvy]

The lim sup is the sup of the set of all subsequential limits. That means that, calling the lim sup a, given any \epsilon> 0 there exist a subsequential limit within \epsilon of a (and less than a). In particular, for every positive integer m, there exist a subsequential limit within 1/2m of a. And because that is a limit of a subsequence, there exist a member of that subsequence, call it a_{m}, within 1/2m of that subsequential limit. Look at what the subsequence a_m converges to. You are forming a new subsequence by, essentially, taking a member from every subsequence.