For any Cauchy sequence in a Banach space we can pick the subsequence $\{{x_{n_k}\}}_{k=1}^\infty: \|x_{n_{k+1}}-x_{n_k}\|\lt 2^{-k} $?

Question

I want to show that given a Cauchy sequence, $\{x_n\}_{n=1}^\infty$ in a Banach space, we can always form a subsequence:

$$\{{x_{n_k}\}}_{k=1}^\infty: \|x_{n_{k+1}}-x_{n_k}\|\lt 2^{-k} $$

What I see is that, since $\{x_n\}_{n=1}^\infty$ is Cauchy, for any $\epsilon$, and $n>m>N(\epsilon)$ we have $\|x_n-x_m\|<\epsilon$. Take $\epsilon = \frac12$ and set $m>n_1>N$ and we obtain $\|x_{m}-x_{n_1}\|\lt 2^{-1}$

We set $m=n_2$ and therefore the above is $\|x_{n_2} - x_{n_1}\|\lt 2^{-1}$

Then we can take $m_2> n_2 >N_2(\epsilon)$ and set $\epsilon = 2^{-2}$ and obtain:

$$\|x_{m_2} - x_{n_2}\|\lt 2^{-2}$$ Setting $m_2 = n_3$ we obtain:

$$\|x_{n_3} - x_{n_2}\|\lt 2^{-2}$$

Using these ${x_{n_k}}$ by induction yields the wanted subsequence, correct?

Answer 1

You're correct in that is how we construct this subsequence, the part about setting $m_p=n_p$ is however superflous as that just yields 0 which is less than any non-zero positive number so it always holds.