Is Rudin being redundant in this proof?

Question

My question is about Definition 3.16 and Theorem 3.17 in Baby Rudin.

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 $x> s^*$, 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).

Of course, an analogous result is true for $s_*$.

Proof (a) If $s^* = +\infty$, then $E$ is not bounded above; hence $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \rightarrow +\infty$.

If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28. [I'll state these theorems after the proof. ]

If $s^* = -\infty$, then $E$ contains only one element, namely $-\infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n \rightarrow -\infty$.

This establishes (a) in all cases.

(b) Suppose there is a number $x > s^*$ such that $s_n \geq x$ for infinitely many values of $n$. In that case, there is a number $y \in E$ such that $y \geq x > s^*$, contradicting the definition of $s^*$.

Thus $s^*$ satisfies both (a) and (b).

To show the uniqueness, suppose there are two numbers, $p$ and $q$, which satisfy (a) and (b), and suppose $p < q$. Choose $x$ such that $p < x < q$. Since $p$ satisfies (b), we have $s_n < x$ for $n \geq N$. But then $q$ cannot satisfy (a).

My question is about this paragraph:

If $s^* = -\infty$, then $E$ contains only one element, namely $-\infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n \rightarrow -\infty$.

I understand the paragraph, but what I do not understand is why the second sentence is necessary.

The way I see it, is that since we assume $s^*$ exists, $E$ must be non-empty. But then, since $s^* = sup E = - \infty$, $E$ cannot possibly contain other elements than $- \infty$ (this would lead to contradiction) and since it contains at least one element, $E = \{ -\infty \}$. Hence $s^* \in E$.

Isn't this enough to prove that $s^* \in E$? Why does Rudin proceed to show $s^* \in E$ by the definition of $E$; this isn't necessary right?

If this were written in a different textbook, I would probably not ask this question, but since Rudin is not known for beating around the bush in his proofs, I'm wondering if I'm missing something.

Thanks for your efforts!

Answer 1

The set $E$ cannot be empty.

Consider the sequence $t_n=\arctan s_n$. Then this sequence has values in $[-\pi/2,\pi/2]$. Thus, by Bolzano-Weierstrass, it has a convergent subsequence; if this subsequence is $(t_{n_k})$, then there are three cases:

1. the subsequence converges to a point $l\in(-\pi/2,\pi/2)$; then $s_{n_k}\to\tan l$;
2. the subsequence converges to $\pi/2$; then $s_{n_k}\to\infty$;
3. the subsequence converges to $-\pi/2$; then $s_{n_k}\to-\infty$.

I don't know whether at this stage the Bolzano-Weierstrass theorem is already known. If it isn't, then the fact that $s^*=\sup E=-\infty$ tells you that either $E=\{-\infty\}$ or $E=\emptyset$.

In the first case, if a subsequence has a limit, then it is $-\infty$ and one such subsequence exists. In the second case, no subsequence has a limit. In either case, no subsequence has a finite limit, therefore $\lim_{n\to\infty}s_n=-\infty$ as proved in the second sentence.