# Is Rudin being redundant in this proof?

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).

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.

• It's not very clear; if $E$ were empty, then you would still have that its supremum is $-\infty$ (it is an upper bound, and is the smallest of all upper bounds). So you still need to prove that $-\infty$ lies in $E$; that is, that there is in fact a subsequence whose limit is $-\infty$. The first sentence should be "contains at most one element", not "contains only one element", I think. (The empty set has a supremum, so assuming $s^*$ exists does not imply that $E$ is nonempty). Commented Dec 14, 2018 at 20:47
• I guess that the intended meaning is “If $s^∗=-\infty$, then $E$ contains only one element, namely $-\infty$, and there is no other subsequential limit.” By itself this doesn't guarantee that $\lim_{n\to\infty}s_n=-\infty$ and the sense of the second sentence is exactly proving the limit exists. Commented Dec 14, 2018 at 21:17
• @StevenWagter The set $E$ contains all subsequential limits and possibly also $\infty$ and $-\infty$: it is not the set of all (finite) subsequential limits and this is clearly written out in the definition of $E$. Commented Dec 14, 2018 at 21:24
• Yes, I understood that. I think I now get it; in the case $s^* = - \infty$, we are not sure yet that $E$ is non-empty; though we know it contains at most one element, $-\infty$. Then Rudin goes on to show that $-\infty$ is indeed an element of $E$ by $E$'s definition. Thank both you guys. Commented Dec 14, 2018 at 21:29

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.