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Give an example of a sequence in $\mathbb{R}$ which has no subsequence which is a Cauchy sequence.

I can find out a sequence that is not a Cauchy sequence such as $\{\ln(n)\}$ once $|\ln(n)-\ln(n+1)|=0$ but $|\ln(n)-\ln(2n)|=|\ln(\frac{1}{2})|>\epsilon$
$\forall \epsilon<\ln(\frac{1}{2})$

I can still find a subsequence of the type ${\ln(2n)}_{2n\in\mathbb{N}}$ such that $|\ln(2n)-\ln(2n+1)|=0$

Question:

What should I do to get a sequence that has no Cauchy subsequence?

Thanks in advance!

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    $\begingroup$ In the first place, you should only be looking at sequences which are not convergent, right? $\endgroup$
    – MPW
    Commented Nov 22, 2018 at 22:57
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    $\begingroup$ I do not understand your question. What do you mean by $|\ln(2n)-\ln(2n+1)|=0$? $\endgroup$
    – Carsten S
    Commented Nov 23, 2018 at 10:29

6 Answers 6

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Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|\geq 1$. So, it has no Cauchy sub-sequence.

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Take a sequence $(a_n)$ such that $a_n\to \infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}\to \infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.

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Take the sequence $1,2,3,4,\ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.

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As Foobaz John says, any sequence with $a_n\to\infty$ works, and it's easy to see that $|a_n|\to\infty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|\not\to\infty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.

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    $\begingroup$ Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts. $\endgroup$ Commented Nov 23, 2018 at 18:32
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I didn't understand your analysis but sequence $a_n = \ln n$ doesn't have Cauchy subsequence since its limit is $\infty$

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  • $\begingroup$ I guess this one uses the theorem that every Cauchy sequence converges $\endgroup$ Commented Nov 23, 2018 at 21:15
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Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.

Fact: Every convergent sequence is bounded.

Strategy: Try an unbounded sequence.

Guess: $a_n=n$

Conclusion: (I leave it to you)

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