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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.

Rudin didn't prove that $E \neq \emptyset$.

Why?

Rudin wrote "If $s^* = -\infty$, then $E$ contains only one element" in the following proof.

But, if $E = \emptyset$, then $s^* = -\infty$ and $E$ contains no element.

So, I think Rudin needs to prove that $E \neq \emptyset$.

I cannot understand Rudin's proof.

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.

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    $\begingroup$ $E$ is nonempty. Anyway, the existence of $s^*$ is contingent on the non-emptiness of $E$ so I think it's fair to say that $s^*=-\infty$ implies $E=\{-\infty\}$. $\endgroup$ Jan 19, 2019 at 10:17
  • $\begingroup$ Thank you very much, Lord Shark the Unknown. $\endgroup$
    – tchappy ha
    Jan 20, 2019 at 2:47

2 Answers 2

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Every sequence in $\overline{\mathbb{R}}$ has a convergent subsequence.

If the sequence is bounded, this is trivial by Bolzano's theorem.

Otherwise, the sequence is unbounded. If it is unbounded above, you can find a subsequence that converges to $+ \infty$. If it is unbounded below, you can find a subsequence that converges to $-\infty$.

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  • $\begingroup$ Thank you very much, Math_QED. $\endgroup$
    – tchappy ha
    Jan 20, 2019 at 2:46
  • $\begingroup$ Glad to help you! $\endgroup$
    – user370967
    Jan 20, 2019 at 8:14
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Note that $s^* = \sup E$. So if $s^* =-\infty$, then $E$ cannot contain any other $x \in \mathbb R \cup \{+ \infty\}$, otherwise $s^*$ will be greater than or equal to that element, and hence greater than $-\infty$, which is not possible. So $E = \{-\infty\}$, since $s^*$ exists only when $E$ is non empty.

If you want an argument as to why $E$ is non-empty, note that every sequence has a monotone subsequence, and every monotone sequence certainly has a limit within the extended real line(via taking the (extended real)supremum/infimum of the monotone sequence as a set depending upon whether it is increasing/decreasing), so such a limit is a member of $E$.

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  • $\begingroup$ Thank you very much, астон вілла олоф мэллбэрг. $\endgroup$
    – tchappy ha
    Jan 20, 2019 at 2:46
  • $\begingroup$ You are welcome! $\endgroup$ Jan 20, 2019 at 10:06

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