\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?

  • $\begingroup$ 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

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.


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