Prove that $(X,d)$ is complete if and only if $\forall(x_n) \in X $ satisfying $d(x_n,x_{n+1})<\frac{1}{2^n} \; \forall n \in N$ is convergent in $X$

Question

Prove that $(X,d)$ is complete if and only if $\forall(x_n) \in X $ satisfying $d(x_n,x_{n+1})<\frac{1}{2^n} \; \forall n\in N$ is convergent in $X$

Note that if $(x_n)$ satisfies $d(x_n,x_{n+1})<\frac{1}{2^n} \; \forall n\in N$ then for $m>n$

$d(x_n,x_m)\leq d(x_n,x_{n+1})+ \ldots + d(x_{m-1},x_m)< \frac{1}{2^n}+\ldots + \frac{1}{2^{m-1}} \to 0$ as $n\to \infty$

so $(x_n)$ becomes Cauchy.

Also if $(x_n)$ is Cauchy, then $d(x_n,x_{n+1}) \to 0$ as $n\to \infty$. How can I show that it satisfies the give condition ? Problem is that it must satisfy for all $n\in N$

Answer 1

First, the property of being a complete metric space means that every Cauchy sequence is convergent, then it is not enough to conclude that you have a Cauchy sequence, as you did. In a complete metric space, if $(x_{n})$ satisfies the condition that $\forall n \in \mathbb{N}$, $d(x_{n}, x_{n+1}) < \frac{1}{2^{n}}$ then $(x_{n})$ is a Cauchy sequence (as in your attempt), hence convergent . On the other hand, if you get $(x_{n})$ a Cauchy sequence in $X$, you can apply the definition of Cauchy sequence with $\epsilon = \frac{1}{2}, \frac{1}{2^{2}}, ...$ to get a subsequence $(x_{n'})$ such that $\forall n' \in \mathbb{N}$, $d(x_{n'}, x_{n'+1}) < \frac{1}{2^{n'}}$, then this subsequence is convergent and so the sequence $(x_{n})$ is.