Two definitions of Limit Superior are equivalent

Question

The following is related to two equivalent definitions of limit superior of a sequence.

Let $\{x_n|n \in \mathbb{N}\}$ be a sequence of real numbers.

Let $A \in [-\infty, \infty]$ be the supremum of the set of limits of all convergent subsequences of $\{x_n\}$ (allowing for convergences to $\pm \infty$).

Let $ S\subset [-\infty, \infty]$ consisting of all $x $ such that $x_n \lt x$ for all but finitely many $n$.

Let $B = inf\{x|x \in S\}$.

Prove that $A \geq B$. I have successfully shown $A \leq B$, so just need the $\geq$ part to eventually conclude $A = B$.

I have thought about this question for over 5 hours and have no idea how I should proceed. Along the way of thinking, I got even more further side questions:

1. Is $limsup$ defined only when a sequence is bounded?
2. What if, for a sequence, there is no subsequence with a limit? Do I get $A = -\infty $?
3. For this question, is it necessary to prove by cases? Like when the sequence $x_n$ is bounded or not. Because I know every bounded sequence has a convergent subsequence. Not sure if this is useful.

Could anyone help me with this question and also the side questions? Thanks a million!

Answer 1

Take $M>A$. I shall prove that $M>B$ and this will prove that $A\geqslant B$.

Take $M'\in(A,M)$. Then there are only finitely many $x_n$'s greater than or equal to $M'$; otherwise, you would be able to find a convergent subsequence of $(x_n)_{n\in\mathbb N}$ such that each of its terms is greater than or equal to $M'$. And if $L$ is the limit of such a subsequence, we would have $L\geqslant M'>A$. This is impossible, by the definition of $A$.

Then, by the definition of $S$, no element of $S$ is greater than $M'$ and therefore $B=\sup S\leqslant M'<M$.

Now, concerning your questions:

1. No. Since you also consider the possibility that a sequence converges to $\pm\infty$, $\limsup_nx_n$ always exists.
2. There is no such sequence (again, because you also consider the possibility that a sequence converges to $\pm\infty$).
3. No, it is not.