About subsequence of a Cauchy sequence

Question

I have a doubt about Cauchy sequence. In a proof I was reading, it is said “Since $\ (u_n) $ is a Cauchy sequence, we can choose an increasing $\ (n_k) $ in $\ \mathbb{N} $ so that $\ || [u_(n_(k+1)) - u_(n_k)] || < 1/(2^k) $. How can it be explained? Thanks

Answer 1

There is some $n_1\in\mathbb N$ such that $m,n\geqslant n_1\implies\lVert u_m-u_n\rVert<\frac12$. And there is a $n_2\in\mathbb N$ such that $m,n\geqslant n_2\implies\lVert u_m-u_n\rVert<\frac1{2^2}$. You can assume without loss of generality that $n_2>n_1$. Then you take a $n_3>n_2$ such that $m,n\geqslant n_2\implies\lVert u_m-u_n\rVert<\frac1{2^3}$. And so on…

Answer 2

There exists $m_k$ such that $|u_n-u_m| <\frac 1 {2^{k}} $ if $n , m \geq m_k$. Let $n_1=m_1$ and inductively define $n_k$ as the maximum of $m_1,m_2,...m_k$ and $n_{k-1}+1$. Then $|u_{n_k}-u_{n_{k+1}}|<\frac 1 {2^{k}} $ for all $k$.

Answer 3

$u_n$ is Cauchy hence convergent.

Let $u$ be the limit.

There exists an increasing subsequence

$u_{n_k}$ s.t. $|u_{n_k} -u| <\dfrac{1}{2^{(k+1)}}$;

With

$|u_{n_{k+1}}-u|<\dfrac{1}{2^{(k+2)}}$ we get

$|u_{n_{k+1}} -u_{n_k}| \le$

$|u_{n_{k+1}}-u| +|u _{n_k}-u| <$

$\dfrac{1}{2^{(k+2)}}+\dfrac{1}{2^{(k+1)}}= \dfrac{3}{2^{(k+2)}} < \dfrac{1}{2^k}$.

Recall:

Let $u_n$ be a sequence and

$n_1 < n_2 < n_3 <..$

an increasing sequence of positive integers.

The the sequence

$u_{n_k}$, $k=1,2,3...$ is called a subsequence of $u_n$.