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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.

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 $s_n \to -\infty$.

This 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.

Does this mean "$E \cap \mathbb{R}$ is not bounded in $\mathbb{R}$"?

Then, for example,
Let $\{s_n\}$ be a sequence such that $s_n = n$.
Then $E = \{+\infty\}$ and $s^* = +\infty$.
And $E \cap \mathbb{R} = \emptyset$.
And $\emptyset$ is bounded above.

I am very confused.

I wanna change the above proof to the following proof:

(a)
if $s^* = +\infty$, then $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \to +\infty$.

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    $\begingroup$ Bounded above usually just means $\leq x$ for a finite $x \in \mathbb{R}$. $\endgroup$ – twnly Jan 20 at 2:56
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    $\begingroup$ @twnly: that would be correct in the usual reals, but in the extended reals $+\infty$ is an upper bound for everything as OP cites from p.12 $\endgroup$ – Ross Millikan Jan 20 at 2:59
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    $\begingroup$ I agree with your change to the proof. I worry about criticizing a classic that has gone through many editions, but I think it is clear you understand what is going on. $\endgroup$ – Ross Millikan Jan 20 at 3:20
  • $\begingroup$ Thank you very much, Ross Millikan. I am greatly relieved at your comment. $\endgroup$ – tchappy ha Jan 20 at 3:25

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