1
$\begingroup$

\Let $\{a_n\}$ be a sequence such that for some $\epsilon >0$, $$ \lvert a_n-a_m\rvert \ge \epsilon$$ for all $n\neq m$. Prove that $a_n$ has no convergent subsequence.

My thoughts on this is to prove that $a_n$is unbounded and therefore it has no convergent subsequence.

First, I am able to prove $a_n$ is not convergent since $a_n$ is not a Cauchy sequence. Then, $\lvert a_n-L+L-a_m\rvert=\lvert a_n-L\rvert+\lvert a_m-L\rvert \geϵ$, $\lvert a_n-L\rvert\gtϵ/2$; so, $a_n$ is not bounded.

I feel that my proof is not sufficient. Can anyone help me?

$\endgroup$
  • $\begingroup$ There are unbounded sequences with convergent subsequences. Take for example $a_{2k + 1} = 0$, $a_{2k} = 2k$. $\endgroup$ – Thad Janisse Oct 12 '15 at 5:51
3
$\begingroup$

If there is any subsequence $(a_{n_{k}})$ of $(a_{n})$ that converges, then for every $\varepsilon > 0$ there is some $N \geq 1$ such that $|a_{n_{k}} - a_{n_{l}}| < \varepsilon$ for all $k,l \geq N$; this contradicts the hypothesis.

$\endgroup$

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.