# Thread: Family of monotonic strictly increasing functions...

1. ## Family of monotonic strictly increasing functions...

Definitions:

1. For functions f,g:NN, we say fg if there exist nN such that for nm we have f(m)g(m)
2. Family F of functions from N to N is unbounded if for every function g:NN, exist fFsuch that fg 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

2. ## 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)