2
$\begingroup$

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.

I have no idea where to start with this proof.

$\endgroup$
  • $\begingroup$ $[-M,M]$ is compact. $\endgroup$ – copper.hat May 3 '17 at 4:25
  • $\begingroup$ 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. $\endgroup$ – Zain Patel May 3 '17 at 4:26
3
$\begingroup$

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 $\infty$.

$\endgroup$
1
$\begingroup$

So, you can use some standard results from analysis to deduce that, but one could argue they obscure just how simple this question is once you have the right point of view.

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.