transition of the relation among real-valued functions? We say $f$ is over $g$ is for all $x\in S_f^g$, there exists $y\in S_g^f$ such that $x<y$, where
$S_f^g=\{x\in\mathbb{R}\mid f(x)<g(x)\}$. Then I want to prove or disprove if $f$ over $g$ and $g$ over $h$, then $f$ is over $h$.
I have the following proof. Can anyone help me to check if its valid?
$f$ over $g$ means for all $x$ such that $f(x)<g(x)$, there exists an $y>x$ such that $f(y)>g(y)$. Similarly $g$ over $h$ means that for all $t$ such that $g(t)<h(t)$, there exist $s>t$ such that $g(s)>h(s)$
Thus let arbitrary $u$ be such that $f(u)<h(u)$, then by the definition of $g$ is over $h$, there is an $s$ such that $f(s)<h(s)<g(s)$, thus, by the definition of $f$ is over $g$, the result follows.
Is this proof looks valid? I guess the only unconvincing part is that can we really deduce $f(s)<h(s)<g(s)$? thanks!
 A: No, that is not valid. The first thing you needed to work out was the statement you need to show.

*

*For all $u$ with $f(u) < h(u)$, there is a $v$ with $u < v$ and $h(v) < f(v)$.

You start off with an arbitrary $u$ with $f(u) < h(u)$, which is good. But your next step just doesn't follow. The definition of $g$ over $h$ does not mention either $f$ at all, so it certainly does not give you $f(s) < h(s)$. Neither does your assumption give us this, because there is no reason to expect that $s = u$.
A second smaller issue is that the definition of $g$ over $h$ does not even give you directly that there exists an $s$ with $h(s) < g(s)$. But the reason this is small is that a trivial additional argument does show the existence of such an $s$. But there is still no relationship between $s$ and $u$ or $f$.

You need to back off and examine this more. Try to think of examples of functions $f$ and $g$ with $f$ over $g$. What does that say about their behavior? If $f$ is constant, what would $g$ have to look like for $f$ to be over $g$? If $g$ were constant, what would $f$ have to look like? Can $f$ be over $g$ and $g$ be over $f$ at the same time?
