Cfrgtkky

I don't understand the contradiction we have assumed that $ exists xin mathbb{R}: 2^n < x ,forall n in mathbb{N}$ therefor the first convergence must be true but the author said that it must converge to 0.

5.25 Theorem (Archimedean property 1.) Let $mbox{ R}$ be a real field, and let $xinmbox{ R}$. Then there is an integer $ninmbox{ N}$ such that $n>x$.

Proof: Let $xinmbox{ R}$, and suppose (in order to get a contradiction) that there is no $ninmbox{N}$ with $n>x$. Then $xgeq n$ for all $n$. Now $displaystyle { left{ left[ 0,{xover {2^n}}right]right}}$ is a binary search sequence in $mbox{ R}$. Since $xgeq 2^n$, I have $displaystyle { 1leq {xover {2^n}}}$ for all $ninmbox{N}$. We see that $displaystyle { left{left[ 0,{xover {2^n}}right]right}to 1}$, but clearly $displaystyle { left{left[ 0,{xover {2^n}}right]right}to 0}$. Since completeness of $mbox{R}$ implies that a binary search sequence has a unique limit, this yields a contradiction, and proves the theorem. $mbox{ $mid!mid!mid$}$

Source:http://people.reed.edu/~mayer/math112.html/html1/node32.html