# 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:

Proof(Partial):

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\$.