Family of monotonic strictly increasing functions...

Definitions:

- For functions
*f*,*g*:N→N, we say *f*≤∗*g* if there exist *n*∈N such that for *n*≤*m* we have *f*(*m*)≤*g*(*m*)

- Family
*F* of functions from N to N is **unbounded** if for every function *g*:N→N, exist *f*∈*F*such that *f*≤∗*g* isn't holds.

Question:

F is unbounded family of monotonic strictly increasing functions from N to N. Show that for everyg:N→N and infinite set X(subset of N) exists f in F such that g(n)<f(n) for infinite n in X.

I even don't know from where to start...

Thank very much!

NNN

Re: Family of monotonic strictly increasing functions...

Hope this is correct...

First, notice that saying f≤∗g isn't holds means there exists infinite values y0,y1... such that g(yn)<f(yn).

For each n, let x(n) be the smallest x greater than n. the image of n -> x(n) is an infinite set.

we define h like this : h(n)=g(x(n))

We know that there exist f in F such that f(y0)>h(y0); f(y1)>h(y1).... (yi are values in N)

since f(y0)>h(y0) then f(y0)>g(x(y0)) AND we have x(y0)≥y0, plus we know that f is increasing, so f(x(y0))≥f(y0)>g(x(y0)), so we proved that f(x(y0))>g(x(y0)), and we can do the same thing** for all x(yn) wich are infinitely many**

(sry for mistakes english isn't my native language)