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Let me refer you to:

http://www.math.iitb.ac.in/atm/caag1/balwant.pdf

Part (2) line $4$ can someone please explain why $x_{ni}=x_{n(m)i}$ for all $i \leq m$ and $n \geq n(m)$. I know it should follow by the fact that $n(m+1) \geq n(m)$ for every $m$ but I'm confused with this part.

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    Shouldn't you ask this as a comment to that post since the issue seems to be that you don't understand something that was posted there? A completely separate post, one in which you moreover **omit** all the relevant information, is not the way to do it.2012-05-10

1 Answers 1

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We have $x_{n+1}-x_n\in\widehat{M}_m$ for all $n\ge n(m)$, so $x_{n+1,i}=x_{n,i}\text{ for all }i\le m\text{ and }n\ge n(m)\;.\tag{1}$ Suppose that $x_{n(m)+k,i}=x_{n(m),i}$ for all $i\le m$; then $x_{n(m)+k+1,i}=x_{n(m)+k,i}=x_{n(m),i}$ for all $i\le m$, by $(1)$ and the induction hypothesis. Thus, $x_{n(m)+k,i}=x_{n(m),i}$ for all $i\le m$ and $k\ge 0$, which is the same as saying that $x_{n,i}=x_{n(m),i}$ for all $i\le m$ and $n\ge n(m)$.

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    @user10: It's a little clumsier than necessary, but yes, it works. I'd say at the second stage that there is some $N(2)$ such that $x_{n+1}-x_n\in\widehat{M}_2$ for all $n\ge N(2)$, and we set $n(2)=\max\{n(1),N(2)\}$. In general, at stage m>1 there is $N(m)$ such that $x_{n+1}-x_n\in\widehat{M}_m$ for all $n\ge N(m)$, and we set $n(m)=\max\{n(m-1),N(m)\}$. Then there's no need to rename the sequence.2012-05-10