Show that a Cauchy sequence has a fast-Cauchy subsequence

Question

A sequence $\{x_j\}$ is said to be fast-Cauchy if $\sum_1^\infty d(x_j,x_{j+1})<+\infty$. Show that every Cauchy sequence has a fast-Cauchy subsequence.

**My attempt:**Argue by contradiction, suppose there is no such subsequence. Then for any subsequence $\{{x_j}_k\}$, $\sum_1^\infty d({x_j}_k,{x_j}_{k+1})$ diverges. Vaguely recall from an analysis course, if a series diverges, it doesn't not satisfy the Cauchy criterion. That is to say, there exists $\epsilon>0$ such that for all $k$ and all $p\ge1$, $d({x_j}_{k+1},{x_j}_{k+2})+...+d({x_j}_{k+p}+{x_j}_{k+p+1})\ge\epsilon$.

Then I intuitively think that this is a contradiction, since if the original series is Cauchy, then after a large $N$, since $d(x_m,x_n)$ has to be very small, so that their sum should be smaller than given $\epsilon$, which is a contradiction. But I don't know how I can write down the concrete proof?

Answer 1

Just select $x_{n_{k}}$ ($k=1,2,\ldots$) according to whether $d(x_{n_{k+1}},x_{n_{k}})<2^{-k}$. This is always possible since you can take $\epsilon_{k}=2^{-k}$ in the Cauchy Criterion.

Answer 2

Let $\mathcal{Y}=\{y_k\}_{k=0}^\infty$ be your Cauchy sequence. Choose any $0<r<1$. Then, since $\mathcal{Y}$ is Cauchy, for every power $r^k,\,k\in\mathbb{N}$ $\exists n(k)\in\mathbb{N}$ such that $|y_j -y_{j+1}|<r^k=\epsilon_k,\,\forall j>n(k)$. So now

$$0<\sum_{j=0}^p|y_j -y_{j+1}| <\sum_{j=0}^p r^k$$

for any positive $p$. The geometric series on the right converges, whence the claim is proven.