# Understanding part (a) Theorem 3.17 from Baby Rudin

We have a sequence $$\{s_n\}$$ 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 $$\{s_{n_k}\}$$. The set $$E$$ then contains all subsequential limits plus possibly the numbers $$\pm\infty$$. We define, $$s^*=\text{sup}E,$$ $$s_*= \text{inf}E.$$ Part $$(a)$$ of Theorem of 3.17 claims that $$s^* \in E$$. The proof of this as given in Rudin, uses case by case analyis. While I have understood the cases for which $$s^*= + \infty$$ and for some finite $$x$$, I have a query regarding the case where $$s^*= - \infty$$. Rudin argues that if $$s^*= - \infty$$, then $$E$$ contains only one element namely $$- \infty$$. My query is regarding the case where $$E$$ is an empty set. This means there exists no subsequence of $$\{s_n\}$$ which has either a finite convergent limit or $$s_n \rightarrow \pm \infty$$. The empty set has $$\textit{sup} -\infty$$, (Least upper bound of an empty set) so this seems to be a counterexample which shows that $$s^*=-\infty \notin E$$. Could someone please shed some light on this? One way to get out of the problem would be if I could prove that $$E$$ cannot be empty, and the only element it can contain is $$-\infty$$.

It happens that $$E$$ is never empty, and therefore your counterexample doesn't work.

In fact:

• If the sequence is bounded, it has a convegent subsequence.
• If the sequence has no upper bound, it has a subsequence whose limit is $$\infty$$.
• If the sequence has no lower bound, it has a subsequence whose limit is $$-\infty$$.

So, in each case $$E\neq\emptyset$$.