Last time, we ended at discussion of two equivalent definitions of subsequential limit. I hope the Cantor's diagonalization trick was fun. Today, we prove the following results
If $s_n$ converge to $s$, then every subsequence of $s_n$ converges to $s$.
Every sequence has a monotone subsequence.
If there are infinitely many $s_n$, that is 'larger than its tails' ($s_n \geq s_k$ for all $k \geq n$), then we can take the subsequences of such $s_n$, it is monotone decreasing.
Otherwise, you throw away an initial chunk that contains such $s_n$, then you can always build an increasing sequence (nothing can stop you)
Every bounded sequence has convergent subsequence. (just take a monotone sequence, then it will be convergent)
$\limsup$ and $\liminf$ can be realized as subseq limits.
Let $(s_n)$ be a seq, and $S$ denote the set of all subseq limits. Then, $S$ is non-empty. $\sup S = \limsup s_n$ and $\inf S = \liminf s_n$. $lim s_n$ exists if and only if $S$ contains only one element. (All these are just summary of previously proven results)
$S$ is closed under taking limits. (i.e. $S$ is a closed set)
Proof of this is fun. Suppose one has a sequence of $t_n$ in $S$, and $t_n \to t$. Can we show that $t$ is in $S$ as well? Well, we need to show that for any $\epsilon>0$, there are infinitely many $s_n$ in $(t-\epsilon, t+\epsilon$. First, we find a $t_n$, such that $|t_n - t| < \epsilon / 2$, then we know there are infinitely many $s_m$ with $|t_n - s_m| < \epsilon/2$, thus these same set of $s_m$ will be $\epsilon$-close to $t$.
Discussion time: Ross 11.2, 11.3, 11.5.