Alternative characterization of lim sup or lim inf. Proof explanation

Question

This is a question about Theorem (2.18) in Stromberg.

Let $(x_n)_{n=1}^\infty\subset \mathbb{R}$ and let $a\in\mathbb{R}^\#$ (the $\#$ stands for the extended real numbers). Then (i) $a=\liminf_{n\rightarrow\infty}x_n$ if and only if whenever $\alpha<a$ we have $\lbrace n\in\mathbb{N}:x_n<\alpha\rbrace$ is finite and whenever $a<\beta$ we have $\lbrace n\in\mathbb{N}:x_n<\beta\rbrace$ is infinite;

There is a second part for the $\limsup$, but I'm not interested in it at the moment. My question is about the proof in the only if part. The author's proof goes as follows:

Conversely, suppose that $a$ satisfies both the "$\alpha$ condition" and the "$\beta$ condition" of (i). If $a=-\infty$, then $a\leq y.$

First question: I can see that if $a$ is infinity, $a\leq y$. How does this help to the proof? In other words, is this a proof that for this particular $a$, this is the $\limsup$?

If $a>-\infty$, then, given $\alpha < a$ we can choose $k_0 \in \mathbb{N}$ such that $\alpha \leq x_n$ for all $n\geq k_0$.

Second question: The $\alpha$ condition says that if $\alpha <a$, then there are finitely many elements of the sequence $x_n$ that are less than $\alpha$. How can I be sure that there are infinitely many elements to the right of $\alpha$, why not use before hand the $\beta$ condition?

Answer 1

As suggested by Paramanad Singh, here is the full answer in case someone is interested in it. The question is about the only if part of the theorem, so I will stick to this part alone. I will rewrite the proof given in the book along with comments without edition about the discussion that happened in the comments.

Conversely, suppose that $a$ satisfies both the "$\alpha$ condition" and the "$\beta$ condition". If $a = -\infty$, then $$$a\leq y$. If $a>-\infty$, then, given $\alpha < a$, we can choose $k_0\in\mathbb{N}$ such that $\alpha \leq x_n$ for all $n\geq k_0$.

The $\alpha$ condition means that there are finite elements in the sequence $(x_n)$ smaller than $\alpha$. This implies that there are infinite $x_n$ such that $x_n>\alpha$.

Thus $\alpha< y_{k_0}$ and so we obtain $$\alpha \leq \sup\lbrace y_k:k\in\mathbb{N} \rbrace=y.\quad (1) $$ Since (1) holds for all $\alpha< a$, we have, whether or not $a=-\infty$, that $$a\leq y.\quad (2) $$

Suppose $y<a$. We have seen that $\alpha \leq y$ so $\alpha \leq y < a$. Since reals are dense, there is an $\alpha_1$ such that $y<\alpha_1<a$. But we have proved $\alpha\leq y$ for every $\alpha<a$ so $y<a$ is a contradiction. Thus $a\leq y$.

If $a=\infty$ the proof is complete. If $a<\infty$, let $a<\beta$. By the "$\beta$ condition" we see that $y_k < \beta$ for each $k\in \mathbb{N}$ and therefore $$y\leq \beta.\quad (3)$$ Since (3) holds for every $\beta > a$, we have $$y\leq a.\quad (4)$$

Suppose $\beta <y$. Since $y$ is the supremum, there is $k_0\in\mathbb{N}$ such that $\beta < y_{k_0}$. But then since $y_{k_0} = \inf\lbrace x_n: n\in\mathbb{N}, n\geq k_0 \rbrace$, there are infinite terms in the sequence such that $\beta < x_n$. But this is in contradiction with the "$\beta$ condition" and thus $y\leq \beta$.

Finally combine (2) and (4) to get $a=y=\liminf x_n$.