# Show that if $(x_n)$ is a bounded sequence, then $(x_n)$ converges if and only if $\limsup (x_n) = \liminf (x_n)$

$$\rightarrow$$ $$x_n$$ 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.

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

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 $$L.$$

• 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 $\lim\inf(x_n) = \lim\sup(x_n) = x$ – Evan Kim Mar 1 '20 at 17:05
• 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. – Matematleta Mar 1 '20 at 18:30