This question is from here:
I want to show that $z=\inf B$ is one of the fixed points of $h$.
I have done this:
Proof(Partial):
If $z\in B$, then $h(z)\le z$. If $h(z)<z$, then $h(h(z))\le h(z)$ because $h$ monotonically increasing, so $h(z)\in B$ so $z$ is not lower bound.
If $z\notin B$, then $h(z)>z$. So, $h(h(z))\ge h(z)$. If $h(h(z))=h(z)$, then $h(z)\in B$. Take any $x\in B$. Since $z\notin B$, and $z$ is lower bound of $B$, we see that $z<x$. So, $h(z)\le h(x)$. Because $x\in B$, $h(x)\le x$ and so we get $h(z)\le x$ for all $x\in B$, so $h(z)$ is lower bound of $B$ greater than $z$, contrary to $z$ being infimum.
What I can't prove is : If $z\notin B$, then $h(z)>z$. So, $h(h(z))\ge h(z)$. If $h(h(z))>h(z)$, then how to come up with a contradiction?
Assume $h(z) > z$ and put $a := h(z)-z$. Then for all $\varepsilon\in (0,a)$ we have $$ h(z+\varepsilon)\,\ge\,h(z) = z+a > z+\varepsilon. $$ Hence, $z+\varepsilon\notin B$ for all $\varepsilon\in (0,a)$. But this means that $z$ cannot be the infimum of $B$.