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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
    Commented Jan 20, 2019 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 $\endgroup$ Commented Jan 20, 2019 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. $\endgroup$ Commented Jan 20, 2019 at 3:20
  • $\begingroup$ Thank you very much, Ross Millikan. I am greatly relieved at your comment. $\endgroup$
    – tchappy ha
    Commented Jan 20, 2019 at 3:25

1 Answer 1

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Short answer: I think you are slightly unclear on the definition of an upper limit. Basically, though, if $\sup(E) = +\infty$, then $E \cap\mathbb{R}$ is not bounded above (in $\mathbb{R}$). We use this to show that $\{s_n\}$ is not bounded above (in $\mathbb{R}$, recall $\{s_n\}$ is a sequence in $\mathbb{R}$, not $\mathbb{R} \cup\{\pm\infty\}$). We use the fact that $\{s_n\}$ is not bounded above, to show that there exists a subsequence which diverges to $+\infty$. So, $+\infty\in\mathbb{R}.$

Long answer:

Let $(a_n)$ be a sequence in $\mathbb{R}.$ Define $A$ to be the subset $A$ of $\mathbb{R} \cup\{\pm\infty\}$ consisting of all subsequential limits of $(a_n)$ and, also, $\pm\infty\in A$ if there exists a subsequence of $(a_n)$ which diverges to $\pm\infty$, respectively.

$a^*$ is merely a shorthand $\limsup_{n\to\infty} a_n$. The definition of $\limsup_{n\to\infty} a_n$ is $\sup(A)$ where $A$ is defined as above. (This is what definition 3.16 means).

Therefore, $\sup(A) = +\infty$ means, by definition of supremum, that if $x\in\mathbb{R} \cup\{\pm\infty\}\ni x<+\infty$ then $x$ is not an upper bound of $A$. In other words, no real number is an upper bound of $A$. In other words, as subset of $\mathbb{R}$, $A$ is not bounded above, by the definition of ''bounded above'' (see definition 1.7). Moreover, $A\cap\mathbb{R}\neq\varnothing,$ since $\sup(\varnothing) = -\infty$ (see here). Therefore, we avoid vacuous statements.

In other words, since $A$ is not bounded above in $\mathbb{R},$ $$ \forall M\in\mathbb{R},\hspace{1mm}\exists\hspace{1mm} a\in A \cap\mathbb{R}\ni a> M. $$ In other words, viewing the definition of $A$, for all $M\in\mathbb{R}$ there exists a real number $a>M$ for which there exists a subsequence of $(a_n)$ which converges to $a$.

Therefore, fix an arbitrary $M \in\mathbb{R}$ and consider the real (i.e., $a \neq +\infty$) $a>M$ for which there exists a sub-sequence $(a_{n_k})$ of $(a_n)$ such that $a_{n_k} \to a$. Consider, also, this sub-sequence $(a_{n_k})$. Using the definition of convergence of a sequence, we have that for the positive quantity $a -M$, there exists an $N\in\mathbb{Z}\ni k \geq N \implies |a - a_{n_k}| <a -M$. Consider any integer $k \geq N$. So, now we have fixed a $k\in\mathbb{Z}$ such that $|a - a_{n_k}| <a -M$. It folows (since $a - a_{n_k} \leq |a - a_{n_k}| < a - M$) that $a_{n_k} > M$. Now, forget that we fixed $M$. This shows that $\forall M \in\mathbb{R}$, some term of $(a_n)$ is greater than $M$. Hence, $(a_n)$ is not bounded above. With this in hand, it is relatively easy to construct a sub-sequence of $(a_n)$ which diverges to $+\infty.$ This, I leave to you.

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