For pseudo-Cauchy sequence, does there always exist a pseudo-Cauchy subsequence of $(x_n)$ having distinct terms?

Question

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:

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?

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!

Answer 1

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.