I have the following problem:
I'm only inspecting sup $A$ = lim inf $s_n$ since I imagine the proof will be more or less the same for both, but I haven't made much progress on this so far.
For the set $S$ of all subsequential limits of $s_n$, I know lim inf $s_n$ = inf $S$, and that the Bolzano-Weierstrass Theorem tells me $s_n$ has a subsequence which converges to some(any?) point in $[0,1]$. I'm not sure if I'm starting to look in the right place and just don't see it yet, or if I'm still completely lost.
I'm not sure if this is correct, or if it matters at all, but for any $a$ in $A$ we have $s_n < a$ for finitely many $n$. Since $s_n$ has infinitely many terms does this imply there are infinitely many terms of $s_n$ greater than $a$?
Thanks for any and all help!