(Baby Rudin, Theorem 3.19)
I am trying to prove:
Let $\{s_n \}$ and $\{t_n \}$ be sequences of real numbers. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then $$\lim_{n\to\infty} \sup s_n \leq \lim_{n\to\infty} \sup t_n.$$
I know this theorem has been proved many times in the past on this website, but it seems like all the proofs that were provided implicitly (and erroneously) assume that both $\lim_{n\to\infty} \sup s_n = s^*$ and $\lim_{n\to\infty} \sup t_n$ are finite. Since this need not necessarily be the case, I thought of asking a new question. Since the finite cases have already been addressed, it remains to deal with the infinite cases:
My attempt at completing the proof:
Suppose $t^* = +\infty$. Then, the result clearly follows; so, assume that $t^* < +\infty$. [Then, I prove that this implies that $s^* < +\infty$]. Now, suppose $s^* = -\infty$ and the result clearly follows; so, assume that $s^* > -\infty$. Then, I need to show that $t^* > -\infty$. When this is done, we can assume that both $s^*, t^*$ are finite.
How can I show that $t^* > -\infty$ in the proof above?
Rudin has the following theorems/definitions related to lim-sup and lim-inf:
Definition 3.15:
Let $\{s_n \}$ be a sequence of real numbers with the following property: For every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \geq M$. We then write $$s_n \rightarrow +\infty.$$ Similarly, if for every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \leq M$, we write $$s_n \rightarrow -\infty.$$
Definition 3.16:
Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \rightarrow x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits as defined in Definition 3.5, plus possibly the numbers $+\infty$, $-\infty$.
We now recall Definitions 1.8 and 1.23 and put $$s^* = \sup E,$$ $$s_* = \inf E.$$ The numbers $s^*$, $s_*$ are called the upper and lower limits of $\{s_n \}$; we use the notation $$\lim_{n\to\infty} \sup s_n = s^*, \ \ \ \lim_{n\to\infty} \inf s_n = s_*.$$
Theorem 3.17:
Let $\{ s_n \}$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:
(a) $s^* \in E$.
(b) If $x > s^*$, then there is an integer $N$ such that $n \geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
Best Answer
From your question, it looks like you're only interested in showing that $t^* > -\infty$ when $s^* > -\infty$.
Since $s^* > -\infty$, there exists a subsquence $s_{n_k}$ such that $s_{n_k} \to s$ for some $s > -\infty$.
Now, consider the subsequence $\{t_{n_k}\}$, it must have a (sub-)subsquence that converges to some $t$. However, $t_{n_k} \ge s_{n_k}$ for all sufficiently large $k$ and thus, $t \ge s > -\infty$.
Since $t^*$ is the supremum of all possible subsequential limits, we see that $t^* \ge t > -\infty$.