"How many orders of infinity are there?" - via MathOverflow

Let F = {f_n(x) = nx} for n E N. For any growth function g in the collection of all growth functions G, there is an integer n such that f_n(x) > g(x) for each x. Proof: else there wouldn't exist natural n > g(x). Of course, for any function that is increasing with anything higher than a multiplicative constant, there exists y > x such that g(y) > f_n(y), but there also exists already m >> n s.t. f_m(y) > g(y), and f_m E F. Then F is cofinal but countable.

Orders of infinity: the 'infinitärcalcül' of Paul Du Bois-Reymond

This is compatible.

## Tuesday, September 28, 2010

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