Thursday, September 30, 2010

I posted this as an answer and while it is a point about the development of the argument, that's not a discussion forum.

"Your set F is not cofinal. None of your f_n dominates g:x->x^2". – Robin Chapman

For each $x E N$, $\exists m >> n$ s.t. $f_m > g(x)$. Let $\mathcal{F}\g(x)$ be $f_n E F: f_n(x) > g(x)$. It is so that F\g(x) is non-empty, for each x E N, for each g_x E G. Via induction, for g(x+1) there exists m > n such that f_m in F\g(x+1). Then, where that is not so, there does not exist m >> n E N.

It's the statement: ! Exists m >> n.

So, if F\g(0) is non-empty, and F\g(k+1) is non-empty for each k, then there doesn't exist x s.t. F\g(x) is non-empty (unless rather induction fails). For each x, F\g(x) is non-empty, for any g.

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