I have the following theorem:
Suppose $a_n$ is a bounded sequence and $u=\limsup(a_n)$. Then:
i) There is a subsequence $a_{n_k}$ with $a_{n_k}\rightarrow u$
ii) If $a_{m_k}$ is convergent subsequence with limit x, then $x\leq u$
Do we have an analogous result for $\liminf (a_n)$? (I'm pretty sure we do but it's not in my notes which seems strange)
Thanks very much for any help
Yes. One can even derive one pair of consequences from the other pair via an inherent symmetry:
$$\limsup (-a_n) =-\liminf a_n$$ $$\liminf (-a_n)=-\limsup a_n$$
For example: Suppose $\lim\inf a_n=u$. Then $\limsup(-a_n)=-u$, thus there is a subsequence of $-a_n$ converging to $-u$ (negate all signs in this subsequence, and that of $-u$, and we have our analogue of result #1). Furthermore if a subsequence of $-a_n$ converges to $-x$, we have $-x\le -u$ by (ii) for the $\limsup$ case, whence $x\ge u$ (notice the negation of the subsequence converges to $x$).