$\rightarrow$ $x_n$ converges implies $\limsup(x_n) = \liminf(x_n)$.
Suppose $(x_n)$ is convergent. Then there exists $x \in \mathbb{R}$ such that every convergent subsequence converges to $x$ in the set $S$ of all subsequential limits of $(x_n)$. I want to show now that $\limsup(x_n) = \liminf(x_n) = x$. To show contradictions that $\limsup(x_n) = x$, I think I need to show that $x< \limsup(x_n)$ and $x > \limsup(x_n)$.
Suppose $x > \limsup(x_n)$. Then this implies that $x \notin S$. However this would contradict the fact that $x$ is a subsequential limit of $x_n$. Suppose that $x < \limsup(x_n)$. Then $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $x < x_n$ for $\epsilon > 0$. However this contradicts the fact that $(x_n)$ is convergent. Similar reasoning can be applied to show $x = \liminf(x_n)$.
$\leftarrow \limsup(x_n) = \liminf(x_n)$ implies $(x_n)$ is convergent
If $\limsup(x_n) = \liminf(x_n)$, this implies that the interval/set $S$ of subsequential limits only has a single value, let's call it $x$. Since we know $(x_n)$ is bounded, we know that there exists a convergent subsequence in $(x_n)$ such that the limit point x' is in $S$. Thus $\limsup(x_n) = \liminf(x_n) = x' = x$. This implies that $\exists N \in \mathbb{N}$ $\forall n \geq N$, that $x + \epsilon < x_n$ for $\epsilon > 0$. Then by observation,
$$ x – \epsilon < x_n \\ x_n – x < \epsilon \\ |x_n – x| < |\epsilon| = \epsilon$$
Since this holds for any $\epsilon > 0$, I conclude that $(x_n)$ is convergent to $x$.
Is my proof solid? Not sure if all of my reasoning makes complete sense to other people.
Best Answer
Maybe it's easier to argue directly, like this:
Set $g_k=\inf_{n\ge k}x_n\ \text{and}\ h_k=\sup_{n\ge k}x_n\tag1$ Now since
$\liminf (x_n):=\underset {k\to \infty}\lim g_k,\ \limsup (x_n):=\underset{k \to \infty}\lim h_k\ \text{and}\ \ g_k\le x_k\le h_k,\tag2$
if $\ \liminf (x_n)=\limsup (x_n),\ \tag3 $
then $(x_n)$ converges by the squeeze theorem.
On the other hand, if $(x_n)\to L$, then there is an integer $N$ such that $L-\epsilon< x_n< L+\epsilon$ whenever $n\ge N$. Then, by definition of $(g_k)$ and $(h_k),$ and because $(g_k)$ is increasing and $(h_k)$ is decreasing, we have, for $n>N,$
$L-\epsilon\le g_N\le g_n\le h_n\le h_N\le L+\epsilon\tag 4$
and this implies that $(g_n)$ and $(h_n)$ converge to $L.$