May I refer you to page 3 of:
http://www.math.iitb.ac.in/atm/caag1/balwant.pdf
Proof that $\hat{M}$ is complete, where it says "We choose $n(m)$ such that $n(m+1) \geq n(m)$ for every $m$".
Question: Why is this possible?
May I refer you to page 3 of:
http://www.math.iitb.ac.in/atm/caag1/balwant.pdf
Proof that $\hat{M}$ is complete, where it says "We choose $n(m)$ such that $n(m+1) \geq n(m)$ for every $m$".
Question: Why is this possible?
Look at the preceding sentence:
For every $m\ge 0$ there exists $n(m)\ge 0$ such that $x_{n+1}-x_n\in\widehat{M}_m$ for all $n\ge n(m)$, ...
In other words, $n(m)$ is being chosen so that something ‘nice’ happens for every $n\ge n(m)$. If some number works for $n(m)$, any larger number also works. If, say, it were the case that $x_{n+1}-x_n\in\widehat{M}_m$ for all $n\ge 100$, then it would also be the case that $x_{n+1}-x_n\in\widehat{M}_m$ for all $n\ge 1000$.
Thus, you can choose the numbers $n(m)$ one at a time. First you choose $n(0)$ to be any number big enough so that $x_{n+1}-x_n\in\widehat{M}_0$ for all $n\ge n(0)$. Then you find a number $k$ such that $x_{n+1}-x_n\in\widehat{M}_1$ for all $n\ge k$, and you let $n(1)=\max\{k,n(0)+1\}$; this guarantees both that $n(1)>n(0)$ and that $x_{n+1}-x_n\in\widehat{M}_1$ for all $n\ge n(1)$. And you continue in this way, inductively choosing $n(m+1)$ to be both big enough to ‘work’ and big enough to exceed $m(n)$.