Let $x_n$ be a sequence of real numbers.
Definition: $x \in \mathbb R \cup \{-\infty,\infty\}$ is a limit point of a sequence $x_n$ if there is a subsequence $x_{n_k}$ of our sequence such that $x_{n_k} \to x$.
If $A= \{x \in \mathbb R \cup \{-\infty,\infty\} \mid x \text { is a limit point of }x_n\}$, prove that $A$ is not empty.
Best Answer
A proof sketch assuming Bolzano-Weierstrass.
Show that $x_n$ is bounded above if and only if $\infty$ is not a limit point of $x_n$.
Reversing the above argument, show that $x_n$ is bounded from below if and only if $- \infty$ is not a limit point.
From (1.) and (2.), conclude that if neither $\infty$ nor $-\infty$ is a limit point of $x_n$, then $x_n$ is bounded. Further, using Bolzano-Weierstrass, conclude that if neither $\infty$ nor $-\infty$ is a limit point of $x_n$, then $x_n$ has a subsequence converging to some limit $x \in \mathbb R$.
To summarize the argument, the key idea is that the sequence is
Note: Martin's answer explains the idea behind (1.). :-)