This is a generalization of [an older question](https://mathoverflow.net/questions/273611/about-the-existence-of-a-convergent-sequence).


Let $(X,d)$ be a metric space and let $(A\_n)\_{n\in\mathbb{N}}\subseteq X$ be a sequence of non-empty closed subsets such that for all $\varepsilon > 0$ there is $N\in\mathbb{N}$ such that for all $m,n\geq N$ we have $d\_H(A\_m,A\_n)<\varepsilon$ (where $d\_H(\cdot,\cdot)$ denotes the [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance)).


Can we pick $x\_n\in A\_n$ for all $n\in\mathbb{N}$ such that $(x\_n)\_{n\in\mathbb{N}}$ is a Cauchy sequence?



Hausdorff distance and Cauchy sequences
Yes. Choosing a subsequence $n\_1<n\_2<\dots$ such that $d\_H(A\_{n\_i},A\_{n\_i+1})\leqslant 2^{-i}$. Then inductively choose points $x\_{n\_i}\in A\_{n\_i}$ so that $d(x\_{n\_i},x\_{n\_{i+1}})\leqslant 2^{-i}$. It is Cauchy sequence by triangle inequality. For any index $m\notin \{n\_1,n\_2,\dots\}$ choose any $i$ such that $n\_i>m$ and a point $x\_m\in A\_m$ on the distance at most $d\_H(A\_m,A\_{n\_i})+1/m$ from $x\_{n\_i}$. The whole sequence $(x\_1,x\_2,\dots)$ is still Cauchy.


