- #1
brian44
- 23
- 0
Hi, I am trying to prove that
[itex]
\{n\}_{n\in \mathbb{N}}
[/itex]
does not converge (based on definition of convergence).
I can prove this by contradiction saying assume it converges, fix [itex] \epsilon [/itex] , then [itex] x_n < \epsilon + a [/itex] (for [itex] n \ge N [/itex] where N is fixed) (by fundamental theorem of ineq.) but by Archimedean Principle, I can find a natural number that surpasses this bound, i.e. [itex] \exists m , m x_n > \epsilon + a [/itex] which is an element of the sequence [itex]x_n[/itex] which means for some M, [itex] n \ge M \rightarrow x_n > \epsilon + a [/itex] which is a contradiction.
However this seems like a long complicated proof for a very simple and obvious fact, I was wondering if there is not some easier, more elegant way to prove this that I am missing?
Thanks for your help.
-Brian
[itex]
\{n\}_{n\in \mathbb{N}}
[/itex]
does not converge (based on definition of convergence).
I can prove this by contradiction saying assume it converges, fix [itex] \epsilon [/itex] , then [itex] x_n < \epsilon + a [/itex] (for [itex] n \ge N [/itex] where N is fixed) (by fundamental theorem of ineq.) but by Archimedean Principle, I can find a natural number that surpasses this bound, i.e. [itex] \exists m , m x_n > \epsilon + a [/itex] which is an element of the sequence [itex]x_n[/itex] which means for some M, [itex] n \ge M \rightarrow x_n > \epsilon + a [/itex] which is a contradiction.
However this seems like a long complicated proof for a very simple and obvious fact, I was wondering if there is not some easier, more elegant way to prove this that I am missing?
Thanks for your help.
-Brian