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$\rightarrow$ $x_n$ 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.

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  • $\begingroup$ you mean $(x_n) $ is a bounded sequence then .. $\endgroup$ – infinity Mar 1 '20 at 13:43
  • $\begingroup$ @infinity whoops yes $\endgroup$ – Evan Kim Mar 1 '20 at 13:44
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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 $L.$

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  • $\begingroup$ Why was it assumed that $g_k$ is increasing and $h_k$ is decreasing? I think for the direction that assumes $(x_n)$ to be convergent, it makes sense that if $(x_n)$ is convergent, then all convergent subsequences in the set of subsequential limits converge to the same value, $x$. Thus $\lim\inf(x_n) = \lim\sup(x_n) = x$ $\endgroup$ – Evan Kim Mar 1 '20 at 17:05
  • $\begingroup$ It is not $assumed$ that they are increasing/decreasing, resp. These facts follow directly from the definition of $g_k$ and $h_k$. As $k$ increases, you are infing/suping over sets, each of which is a subset of the previous one. $\endgroup$ – Matematleta Mar 1 '20 at 18:30

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