Why does every sequence of real numbers have an upper limit?

Question

Let $\{a_n\}:\mathbb{N}\to\mathbb{R}$, and we define $E$ to be the set of all sub-sequential limits of $\{a_n\}$ as well as possibly the symbols $\infty$ and $-\infty$ if there are some sub-sequences of $\{a_n\}$ that go to $\infty$ / $-\infty$.

Can $E$ be empty? In particular, since $\limsup_{n \to \infty}a_n := \sup E$, does the upper limit of a sequence of real numbers is always defined? From what I found on the Wikipedia page on limit superior and limit inferior, the answer is yes and this have something to do with the fact that $\mathbb{R}$ is complete, but I couldn't find the complete proof, and I'm a bit surprised the issue isn't discussed on Rudin's principles of mathematical analysis.

Answer 1

Every sequence of real numbers has a monotone subsequence. This monotone subsequence converges to something in $\mathbb{R} \cup \{-\infty,\infty\}$, i.e. $E \neq \emptyset$.

If you would like to see a proof of this fact, look up the Monotone Subsequence Theorem.

Answer 2

Let $\sigma=\langle a_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R$. If $\sigma$ is unbounded, then it has a subsequence converging to $\infty$ or to $-\infty$, so $E\ne\varnothing$. If $\sigma$ is bounded, then by the Bolzano-Weierstrass theorem it has a convergent subsequence, so again $E\ne\varnothing$.