# Completion of a module is equivalent to Cauchy sequence criterion

Suppose $M$ is an $A$-module where $A$ is a commutative ring with $1$. $M$ has the $I$- adic topology where $I$ is an ideal of $A$. We define the completion of $M$, $\hat M:=\varprojlim M/I^nM$. Then we have a natural map $\psi:M\rightarrow \hat M$. We say $M$ is complete if $\psi$ is an isomorphism. Clearly this is equivalent to say that for every sequence $x_1,x_2,\cdots$ of elements of $M$ satisfying $x_i-x_{i+1}\in I^iM, \forall i$, $\exists$ a unique $x\in M$ such that $x-x_i\in I^iM$ for all i. I need to prove that $M$ is complete if and only if every cauchy sequence in $M$ converges to a unique limit in $M$ ($\{x_i\}$ is Cauchy if and only if for every positive integer $r$ there there is an $n_0$ such that $x_{n+1}-x_n\in I^rM$ for $n>n_0$).

I have proved that if every Cauchy sequence has a unique limit then $M$ is complete. But I cannot prove the other way around, i.e., if $M$ is complete then every Cauchy sequence has a unique limit.

Can anyone help me how to prove this?

Thank you.

I think I have got an answer of the above question. Please point out if there is any mistake:

Suppose $M$ is $I$-adically complete, i.e., for any sequence $x_1,x_2,\cdots\mspace .11in$ of elements in $M$ with $x_n-x_{n+1}\in I^nM, \forall n\in\mathbb N$, there exists a unique $x\in M$ such that $x-x_n\in I^nM,\forall n\in\mathbb N$.

Now let $x_1,x_2,\cdots\mspace .11in$ be a cauchy seuence of elements of $M$. Then $\exists n_1\in\mathbb N$ such that $x_r-x_{r+1}\in IM,\forall r\geq n_1$, in particular $x_{n_1}-x_{n_1+1}\in IM$. Also $\exists n_2\in\mathbb N$ such that$x_r-x_{r+1}\in IM,\forall r\geq n_2$, in particular $x_{n_2}-x_{n_2+1}\in I^2M$. Without loss of generality we can have $n_2>n_1$. Similarly $\exists n_3\in\mathbb N(n_3>n_2)$ such that $x_r-x_{r+1}\in I^3M,\forall r\geq n_3$, in particular $x_{n_3}-x_{n_3+1}\in I^3M$. We continue this process

Now if we take $y_1=x_{n_1}, y_2=x_{n_2}, y_3=x_{n_3},\cdots$

We have:

$$y_1-y_2=x_{n_1}-x_{n_1+1}+x_{n_1+1}-x_{n_1+2}+\cdots +x_{n_2-1}-x_{n_2}\in IM.$$ So in general we have $y_n-y_{n+1}\in I^nM$, for all $n\in\mathbb N$. Now using the above hypothesis we have a unique limit for the subsequence $\{y_n\}_{n\in\mathbb N}$.

So it is enough to prove that if a Cauchy sequence in $M$ have a convergent subsequence converging to a unique limit then the Cauchy sequence itself converges in the above sense to that limit and it is also unique.

Let $\{x_n\}$ be a Cauchy sequence in $M$ and it has convergent subsequence $\{x_{n_k}\}$ converging to a unique limit $x$ in $M$. fix $n\in\mathbb N$.

We have $$x_u-x_{u+1}\in I^nM,\forall u\geq r.$$ Without loss of generality we can have $r>n$. Also choose $n_k\geq r$. Then $$x-x_v\in I^{n_k}M,\forall v\geq n_l.$$ Choose $n_a\geq n_k$ and $n_a\geq n_l$. Then $$x-x_{n_a}\in I^{n_k}M\subseteq I^nM,$$ Now for any $s\geq n_a$, $$x-x_s=x-x_{n_a}+x_{n_a}-x_{n_a+1}+\cdots +x_{s-1}-x_s\in I^nM.$$ Hence $\{x_n\}$ converges to $x$. Now it is easy to see that if a Cauchy sequence converges to a limit then every subsequence of that sequence converges to that limit. Since the subsequence limit is unique, we have $\{x_n\}$ converges to the unique limit $x$.