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 $-\infty$.


1 Answer 1


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 $E\neq\emptyset$.


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