# Walter Rudin "Principles of Mathematical Analysis" Definition 3.16, Theorem 3.17. I cannot understand.

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.

• $E$ 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 $E=\{-\infty\}$. Jan 19, 2019 at 10:17
• Thank you very much, Lord Shark the Unknown. Jan 20, 2019 at 2:47

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 $$-\infty$$.

• Thank you very much, Math_QED. Jan 20, 2019 at 2:46
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 $$E$$.