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