If $B$ is a Boolean algebra, then a mapping $f:B\rightarrow B$ is said to be contractive (or a contraction) if $f(a)+f(b)\leq a+b$ for each $a,b\in B$ where $a+b=(a\wedge b')\vee(a'\wedge b)$ is the sum in the corresponding Boolean ring. Suppose that $B$ is a Boolean algebra such that for each contractive mapping $f:B\rightarrow B$, the maximum $Max_{b\in B}f(b)$ is achieved by some $b\in B$. Then is $B$ necessarily complete? What about when we assume that $B$ is originally $\sigma$-complete?

I am looking for counterexamples, so I would be interested in seeing what kinds of counterexamples one could construct even if a counterexample is found.

Every contractive function on a complete Boolean algebra attains its maximum: If $B$ is complete and $f:B\rightarrow B$ is contractive. Suppose that $c=\bigvee_{b\in B}f(b)$. Then there is a partition $p$ of $c$ such that for each $a\in p$, there is some $b_{a}\in B$ with $a\leq f(b_{a})$. Let $x=\bigvee_{a\in p}b_{a}\wedge a$. Then

$f(x)'\wedge a\geq f(x)'\wedge f(b_{a})\geq f(x)+f(b_{a})\leq x+b_{a}\leq a'$, so

$f(x)'\wedge a=0$, hence $f(x)'\leq a'$, thus $a\leq f(x)$. Therefore $f(x)\geq\bigvee_{a\in p}a=c$. However, I am unsure if this theorem holds for more general classes of Boolean algebras besides complete Boolean algebras.