$\limsup$ and $\liminf$ in Extended Reals Is it true that any sequence $X_n$ converges to $x$ in the extended real number system if and only if $\limsup X_n \le \liminf X_n$?
 A: Strictly speaking, what you wrote is not quite true. Indeed, consider the constant sequence $x_n=0$ and set $x:=1$. Then $\liminf x_n=0=\limsup x_n$, but $x_n\not\to x$. However, it's clear that you meant to ask:

If $(x_n)$ is a sequence of real numbers, then $(x_n)$ converges to an extended real number $x$ iff $\limsup x_n \le \liminf x_n$.

Let $(x_n)$ be a sequence of real numbers.
It's easiest to think of $\liminf x_n$ and $\limsup x_n$ as the infimum and supremum, respectively, of all subsequential limits of $(x_n)$.
If this isn't something you've seen before, it's a good exercise to work through. The wikipedia page may also be helpful.
From this definition, or the others, it should be clear that $\liminf x_n \le \limsup x_n$.
With these two remarks in mind, suppose $(x_n)$ converges to some extended real number $x$.
Then every subsequence of $(x_n)$ must also converge to $x$, so by the above characterization, we have $\liminf x_n=\limsup x_n$.
Conversely suppose $\limsup x_n \le \liminf x_n$. Hence $\liminf x_n=\limsup x_n$. We aim to show that if $x:=\liminf x_n$, then $x_n\to x$. Suppose by contradiction that $x_n\not\to x$. Then there exists a neighborhood $U$ of $x$ for which infinitely many terms of the sequence $(x_n)$ lie outside $U$. This gives a subsequence $(y_n)$ of $(x_n)$ such that $y_n\not\in U$ for every $n$. Then any convergent subsequence of $(y_n)$ will not converge to $x$, and thus contradict the assumption $\liminf x_n=\limsup x_n = x$.
It only remains to see that $(y_n)$ has a convergent subsequence. If it is bounded, then this is given to us by the Bolzano-Weierstrass theorem. On the other hand, if it is unbounded, then it has a subsequence which converges to either $\pm\infty$. This completes the proof.
