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?
Best Answer
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.