I am reading Walter Rudin’s “Principles of Mathematical Analysis”.

There are the following definition and theorem and its proof in this book.

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, plus possibly the numbers $ +\infty$ , $ -\infty$ .

Put $ $ s^* = \sup E,$ $ $ $ s_* = \inf E.$ $

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} \to +\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.

, 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 \to -\infty$ .If $ s^* = -\infty$ , then $ E$ contains only one elementThis establishes (a) in all cases.

I cannot understand the following argument:

(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} \to +\infty$ .

What does “$ E$ is not bounded above” mean?

p.12, Rudin wrote “It is then clear that $ +\infty$ is an upper bound of every subset of the extended real number system”.

And $ E$ is a subset of the extended real number system.