# Prove that for every 2 elements in the set F of all functions from N to N, there's an element in F that's bigger than both

let there be $\ F$ the set of all functions from $\ N \rightarrow N$.

K is a relation on F,

for every f,g$\in$F , (f,g)$\in$K $\leftrightarrow$ for all $\ n\in N$, $\ f(n)\leq g(n)$

Prove that for every two elements in $\ F$, there exist an element that's bigger than both.

in other words, given $\ f,g\in F$, proove that there's $\ h\in F$, that sustains

$\ (g,h)\in K,(f,h)\in K$, $\ h$ is different from $\ f,g$.

remark: h is not a constant element of F, it depends on f,g.

for every $\ f,g\in F$, there's $\ h\in F$, such that $\ h(n)=f(n)+g(n)$
$\rightarrow f(n)\leq h(n), g(n)\leq h(n)\rightarrow (f,h)\in K,(g,h)\in K$
• If your convention is that $N$ doesn't contain $0$, then your proof is fine. If (like me) you consider $0$ to be a natural number, then there's the possibility that one of $f$ and $g$ is identically $0$, so the other one equals $h$, which you didn't want. So modify your construction to say $h(n)=f(n)+g(n)+1$. – Andreas Blass Apr 21 '14 at 20:42