# Show that if $x_n$ has no convergent subsequences then $|x_n| \to \infty$

Suppose that the sequence $x_n$ has no convergent subsequences. Let M $> 0$.

Prove that there exists at most finitely many values of $n$ such that $x_n \in [-M,M].$ Explain why this implies |$x_n$|$\rightarrow$$\infty$ as $n \to \infty$

This is part a of 1.62 from Advanced Calculus by Leonard Richardson.

• $[-M,M]$ is compact. – copper.hat May 3 '17 at 4:25
• Do you have Bolzano at your disposal? If so, if there are infinitely many values such that $x_n \in [-M,M]$ then the sequence is bounded and must have a convergent subsequence. Contradiction. – Zain Patel May 3 '17 at 4:26
We argue by contradiction. Assume that there are infinitely many values of $n$ such that $x_n \in [-M,M]$ then take a subsequence $x_{n_k}$ such that $x_{n_k} \in [-M,M]$ for all $k$. Then $x_{n_k}$ is bounded and must have a convergent subsequence by the Bolzano-Weierstrass theorem. So $x_n$ must have a convergent subsequence. Contradiction.
The second part follows from using the definition of what it means for a sequence to go to $\infty$.
If there are infinitely many terms of the sequence in $[-M,M]$, then it's quite obvious there are infinitely many in either $[-M,0]$ or $[0,M]$. Say it's $[0,M]$ WLOG; then continuing there are infinitely many in either $[0,M/2]$ or $[M/2,M]$, and so on. This defines an infinite nested sequence of intervals that will converge to some number (the intersection of all the intervals), and this number is the limit of a convergent subsequence of the $x_i$ (take one term from each interval). So this shows the first part, and the second part now follows easily :)