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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.

my answer:

for every $\ f,g\in F$, there's $\ h\in F$, such that $\ h(n)=f(n)+g(n)$

$\rightarrow f(n)\leq h(n), g(n)\leq h(n)\rightarrow (f,h)\in K,(g,h)\in K$

is it good?

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    $\begingroup$ I only skimmed through it, but it looks good. This question was asked recently, but the other asker wasn't careful enough to say what 'bigger than' meant, making it hard to answer. $\endgroup$ – Git Gud Apr 21 '14 at 19:24
  • $\begingroup$ Possible duplicate, definitely related: math.stackexchange.com/questions/760889/union-of-functions $\endgroup$ – Asaf Karagila Apr 21 '14 at 19:51
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    $\begingroup$ 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 $h(n)=f(n)+g(n)+1$. $\endgroup$ – Andreas Blass Apr 21 '14 at 20:42
  • $\begingroup$ yes Andreas I just thought about this myself and was about to change my answer, wasn't sure about it though, now I am, thanks! $\endgroup$ – Seth Keno Apr 21 '14 at 21:03

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