$M$ is complete iff by showing given sequence is cauchy .

Question

Prove that metric space $(M,d)$ is complete iff every sequence $(x_n)$ in $M$ satisfying. $$d(x_n,x_{n+1}) < 1/2^n$$ for all $n$ converges to a point of $M$.

Attempt: Reverse implication.

Since,

$$d(x_n,x_{n+1}) < 1/2^n$$

for all $n$ to $N \implies d(x_n,x_{m}) < 1/2^{m-n} = \varepsilon _{m,n} \, $ for all $m>n$ and $ m,n \geq N$ for some natural number $N \implies (x_n)$ is a Cauchy sequence which converges to point in $M \implies M$ is complete .

Is this right?

How to go for forward implication?
Is it reverse steps of reverse implication .

Answer 1

Suppose first that $(M, d)$ is complete. Consider a sequence $(x_n)$ such that for all $n \in \mathbb{N}$, $$d(x_n, x_{n+1}) < \frac{1}{2^n}$$

Let $\varepsilon > 0$. There exists $N \in \mathbb{N}$ such that $$\frac{1}{2^{N-1}} < \varepsilon$$

Let $p > q \geq N$. You have then $$d(x_p, x_q) \leq \sum_{k=q}^{p-1} d( x_{k+1},x_k) \leq \sum_{k=q}^{p-1} \frac{1}{2^{k}} = \frac{1}{2^q} \frac{1- \frac{1}{2^{p-q}}}{1 - \frac{1}{2}} = \frac{1}{2^{q-1}} - \frac{1}{2^{p-1}} \leq \frac{1}{2^{q-1}} \leq \frac{1}{2^{N-1}} \leq \varepsilon$$

Therefore your sequence is a Cauchy sequence, and because $(M,d)$ is complete, it converges.

Now suppose that every sequence $(x_n)$ such that $d(x_n, x_{n+1}) < \frac{1}{2^n}$ for all $n$ converges, and let's prove that $(M, d)$ is complete. Consider a Cauchy sequence $(y_n)$. By definition, for all $k \in \mathbb{N}$, there exists $N_k$ such that for all $p, q \geq N_k$, $$d(y_p,y_q) < \frac{1}{2^k}$$ In particular, the sequence $(N_k)$ can be choosen increasing, so you get that for all $k\in \mathbb{N}$, $$d(y_{N_k},y_{N_{k+1}}) < \frac{1}{2^k}$$

Define now the sequence $x_k = y_{N_k}$. You have $d(x_k, x_{k+1}) < \frac{1}{2^k}$ for all $k$, so by assumption, $(x_k)$ converges. Let $l$ be the limit of $(x_k)$. Let's prove that $l$ is also the limit of $(y_n)$.

Let $\varepsilon > 0$. Because $(x_k)$ converges to $l$, there exists $N \in \mathbb{N}$ such that for all $k \geq N$, $$d(y_{N_k},l) =d(x_k, l) < \frac{\varepsilon}{2}$$

Moreover, there exists $N'$ such that $\frac{1}{2^{N'}} < \frac{\varepsilon}{2}$ : by definition, for all $k \geq N'$, for all $n \geq N_k$, you have that $$d(y_n, y_{N_k}) < \frac{\varepsilon}{2}$$

Let's define $N'' = N_{\max{N,N'}}$ : for all $n \geq N''$, you have then that $$d(y_n, l) \leq d(y_n, y_{N_k}) + d(l, y_{N_k}) < \varepsilon$$

So $(y_n)$ converges to $l$. That is sufficient to ensure that $(M,d)$ is complete.

Answer 2

Hints:

For the reverse implication, suppose that $(x_n)_n$ is a Cauchy sequence in $M$. Then inductively construct a subsequence $(x_{p(n)})_n$ such that $d(x_{p(n)}, x_{p(n+1)}) < \frac1{2^n}$. Hence $(x_{p(n)})_n$ converges to some $x \in M$. Now show that $x_n \to x$.

For the forward implication, assume that $M$ is complete and that a sequence $(x_n)_n$ satisfies $d(x_{n+1}, x_n) < \frac1{2^n}$. Then for $m > n$ we have \begin{align} d(x_m, x_n) &\le d(x_{m}, x_{m-1}) + d(x_{m-1}, x_{m-2}) + \cdots + d(x_{n+1}, x_n)\\ &\le \frac1{2^{m-1}} + \frac1{2^{m-2}}+\cdots + \frac1{2^n}\\ &= \frac1{2^n}\left(1+\frac12 +\cdots + \frac1{2^{m-n-1}}\right)\\ &= 2\left(\frac1{2^n} - \frac1{2^m}\right) \end{align} From here conclude that $(x_n)_n$ is Cauchy and hence convergent.