TITLE: Does every sequence have at least one limit point?
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: If the sequence is bounded, then you can invoke Bolzano-Weierstrass theorem.
So we can assume it is unbounded. Let us assume it is not bounded from above. (The symmetric case is similar.)
Thus for every $C$ there exists $n$ such that $x_n>C$.
Start with $C_1=1$ and choose $n_1$ such that $x_{n_1}>C_1=1$.
Now choose $C_2=2+\max\{x_j\; : \; j\le n_1\}$. Obviously $C_2\ge 2$ and you can choose $n_2$ such that $n_2>n_1$ and $x_{n_2}>C_2$. 
You continue by induction. In the $k$-th step you choose $C_k=k+\max\{x_j \; : \; j\le n_k\}$. This means that $C_k\ge k$ and you there exists $n_{k+1}>n_k$ with $x_{n_k}>C_k$.
In this way you obtain a subsequence $n_k$ such that $x_{n_k} \ge k$. This implies that this subsequence converges to $+\infty$.