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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.

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Assume that $B$ is an arbitrary Boolean algebra and that $f\colon B\to B$ is an arbitrary contraction. I will argue that $f$ assumes its maximum at $a=f(1)\wedge f(0)'$, and the maximum value is $f(a)=f(0)\vee f(1)$.

Set $F(x) = f(x) + f(0)$. Then $$F(a)+F(b) = f(a)+f(0)+f(b)+f(0)=f(a)+f(b)\leq a+b,$$ so $F\colon B\to B$ is another contraction. Moreover, this contraction satisfies $F(0)=f(0)+f(0) = 0$. Using this, and setting $b=0$ in $F(a)+F(b)\leq a+b$, yields $F(a)\leq a$.

Now set $b=1$ in $F(a)+F(b)\leq a+b$ to obtain $$ F(a)+F(1)\leq a+1 = a'. $$ If you meet both sides with $a$ you get $$F(a)\wedge a + F(1)\wedge a = 0.$$ But since $F(a)\leq a$, this reduces to $F(a) + F(1)\wedge a = 0,$ or (adding $F(1)\wedge a$ to both sides) $$ F(a) = F(1)\wedge a. $$ Reverting back to little $f$, we get $f(a)+f(0)=(f(1)+f(0))\wedge a$, or $$ f(a) = (f(1)+f(0))\wedge a + f(0). $$ So contractions are entirely determined by the choices of $f(0)$ and $f(1)$. (Moreover, the values of $f$ at $0$ and $1$ can be chosen freely.)

This characterization of contractions shows that any contraction on any BA assumes its maximum at $a=f(1)\wedge f(0)'$, and the maximum value is $f(a)=f(0)\vee f(1)$.

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