Given a metric space $(X,d),$ let us call a sequence $(x\_n)$ in $X$ to be **pseudo-Cauchy** if $\lim\limits\_{n\to\infty}d(x\_n,x\_{n+1})=0.$


For example, the sequence $\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\_n$ is pseudo-Cauchy in $\mathbb R$ without being Cauchy.


*Now I would like to ask the following question:*



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> **Given a pseudo-Cauchy sequence $(x\_n)$ in a metric space $(X,d)$ *having no constant subsequence*, does there always exist a pseudo-Cauchy subsequence of $(x\_n)$ having distinct terms?**
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Intuitively it is appearing to me that there should not always exist such a subsequence; however I failed to construct a counterexample or give a proof.


Please help me!



For pseudo-Cauchy sequence, does there always exist a pseudo-Cauchy subsequence of $(x_n)$ having distinct terms?
The answer to the question is yes. I struggled for a while trying to find the right growth rate so that terms in sequences are sufficiently spread but also not too much, until I realized that the following basic trick is more suited to the problem:


The hypothesis of having no constant subsequence means that each value of the sequence occurs a last time in the sequence. Start with the value $x\_0$ and set $\varphi(0)$ as the largest index with $x\_0=x\_{\varphi(0)}$. Having defined $\varphi(n)$, define $\varphi(n+1)$ as the largest index with $x\_{\varphi(n)+1}=x\_{\varphi(n+1)}$. This insures at the same time that $x\_{\varphi(n+1)}$ is distinct from all $x\_{\varphi(k)},k\leq n$, and that two consecutive terms of the subsequence $(x\_{\varphi(k)})\_{k \in \mathbb{N}}$ are consecutive terms of $(x\_k)\_{k \in \mathbb{N}}$ in an orderly way. Hence $\varphi$ is a solution.


