Prove that a sequence $(x_n)$ is Cauchy given $x_n$ is a point in $E_n$, where each $E_n$ is a nonempty bounded sequence of $\mathbb{R}^d$

Question

Suppose that $E_1 \supseteq E_2 \supseteq ....$, where each $E_n$ is a nonempty bounded subset of $\mathbb{R}^d$, and suppose also that $\lim_{n\to\infty} \mbox{diam}(E_n) =0$. For each $n \in \mathbb{N}$, let $x_n$ be one point in $E_n$. Prove that $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}^d$.

I am not sure at all how to do this. I know a sequence is Cauchy if for every $\epsilon >0$, there exists an $N>0$ such that $m,n \ge N$ implies $|x_m-x_n|\le \epsilon$.

Answer 1

For each $m, n, k\in\Bbb N$ with $m, n\geqslant k$, we have from our hypotheses that $x_m, x_n \in E_k$. This implies that $\lVert x_m - x_n\rVert \leqslant \mbox{diam}(E_k)$. Can you use the fact that $\lim_{n\to\infty} \mbox{diam}(E_n) =0$ to conclude?