How to show $inf B$ is fixed point? Code Answer

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I want to show that $z=inf B$ is one of the fixed points of $h$.

I have done this:


If $zin B$, 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 $znotin B$, then $h(z)>z$. So, $h(h(z))ge h(z)$. If $h(h(z))=h(z)$, then $h(z)in B$. Take any $xin B$. Since $znotin B$, and $z$ is lower bound of $B$, we see that $z<x$. So, $h(z)le h(x)$. Because $xin B$, $h(x)le x$ and so we get $h(z)le x$ for all $xin B$, so $h(z)$ is lower bound of $B$ greater than $z$, contrary to $z$ being infimum.

What I can’t prove is : If $znotin B$, 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 $varepsilonin (0,a)$ we have
h(z+varepsilon),ge,h(z) = z+a > z+varepsilon.
Hence, $z+varepsilonnotin B$ for all $varepsilonin (0,a)$. But this means that $z$ cannot be the infimum of $B$.

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