# Proving that $s_n \le t_n \implies s^* \le t^*$ (Baby Rudin, Theorem 3.19)

(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).

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 $$t^* \ge t > -\infty$$.

• Why would $\{t_n_k\}$ have a subsequence that converges to some $t$? $\{t_n_k\}$ need not be bounded, right (not sure why the equations are being boxed automatically in the comment)? Commented Jun 30, 2020 at 16:05
• Given any sequence, it has a convergent subsequence in the extended sense. If it's unbounded, you can find one that converges to either $\infty$ or $-\infty$. In this case, $-\infty$ won't be possible due to the lower bound by $s_{n_k}$. Commented Jun 30, 2020 at 16:07
• A quick way to see that: Assume wlog that $a_n$ is unbounded above. You can construct a strictly increasing sequence $n_1, n_2, \ldots$ such that $a_{n_1} > 1$, $a_{n_2} > 2$, and so on. Clearly $\lim_k a_{n_k} = \infty$. (The reason you can construct it will follow quite straight-fowardly from the definition of not being bounded above.) Commented Jun 30, 2020 at 16:11
• Also, about the boxing, I think it's because you have to put nested subscripts via brackets. Try \{t_{n_k}\} instead to get $\{t_{n_k}\}$. Commented Jun 30, 2020 at 16:20
• Thanks, I really like the simplicity of your proof! Commented Jun 30, 2020 at 19:31

You could show that $$t_* > -\infty$$, or you could simply show that $$s_* \le t_*$$ by working with the assumption that $$t_* < \infty$$. This will be just as hard (or as easy) as showing that $$t_* > - \infty$$ at this point in my opinion. As you already noted, $$s_* < \infty$$ in this case.

Assume for the sake of contradiction that $$t_* < s_*$$ and let $$s_{n_k}\to s_*$$. By assumption, $$s_{n_k} \le t_{n_k}$$ for all $$k\ge K_1$$ for some $$K_1$$.

If $$t_* = -\infty$$, then for some $$K_2$$ and all $$k\ge K_2$$, $$t_{n_k} < s_* - 1$$. This is impossible because $$s_{n_k}\to s_*$$.

If $$t_* > -\infty$$, so $$t_*\in\mathbb R$$, then $$s_* - t_* > c > 0$$ for some $$c$$. So for some $$K_3$$ and all $$k\ge K_3$$, $$s_{n_k} \le t_{n_k} < t_* + c < s_*$$, another contradiction because $$s_{n_k}\to s_*$$.