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According to Kolmogorov ("Algèbres de Boole métriques complètes", VI Zjazd Matematykòw Polskich, 1948, english translation Phil. Studies, 1995, 77, 57-66), a Boolean algebra $(B, \wedge, \vee,-,1,0)$ is metric if it is given together with a map $ B\xrightarrow\mu R $ to the reals such that:

1) $[x \wedge y=0] \Rightarrow [\mu(x \vee y)=\mu (x) + \mu (y)]$

2) $ [x \neq 0] \Rightarrow [\mu (x)>0] $.

In facts, if $B \times B \xrightarrow+ B$ denotes symmetric difference, the composite $ B \times B \xrightarrow+ B \xrightarrow\mu R$ is a metric.

Denoting: Bool the category of Boolean algebras, Met the category of metric spaces with non-expansive maps and MBool the category of metric Boolean algebras with non-expansive Boolean homomorphisms:

  • what are the main differences between the categories MBool and Bool ?

  • the two forgetful functors MBool**$\rightarrow$ *Bool and MBool*$\rightarrow$ **Met have left adjoints ?

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1 Answer 1

I have a feeling that uncountable products in $MBool$ do not exist in general. One thing is for sure: if $MBool$ has uncountable products, then the forgetful functor $MBool \to Bool$ cannot preserve them (whence the forgetful functor couldn't have a left adjoint.)

Indeed, if $2$ is the 2-element Boolean algebra with its unique measure $\mu$. If the uncountable product $(2, \mu)^{\omega_1}$ existed in $MBool$, where $\omega_1$ is the first uncountable ordinal, and if this product is calculated as it would be in $Bool$ (which is the set-theoretic infinite product), then we could construct an uncountable strictly decreasing chain of elements (uncountable tuples)

$$(1, 1, 1, \ldots) > (0, 1, 1, \ldots) > (0, 0, 1, \ldots) > \ldots$$

and there would have to be a corresponding uncountable strictly decreasing chain of positive real numbers

$$\mu(1, 1, 1, \ldots) > \mu(0, 1, 1, \ldots) > \mu(0, 0, 1, \ldots) > \ldots$$

by the axioms on metric Boolean algebras. This is impossible by cofinality considerations.

Regardless of whether there are infinite products in $MBool$, there is a trivial reason why the forgetful functor $U: MBool \to Bool$ cannot have a left adjoint $F$ under the axioms given for metric Boolean algebras. Suppose WLOG that $\mu(1) = 1$ where the $1$ on the left is the top element of the putative free metric Boolean algebra $FB$. Then define a different metric $\mu'$ on $UFB$ by $\mu'(b) = r\mu(b)$ where $r > 1$, giving a different metric Boolean algebra $B'$. Then there is no nonexpansive metric Boolean algebra map $FB \to B'$ which extends the Boolean algebra embedding $i: B \to UB'$ along $i: B \to UFB$.

One might think this trivial objection could be remedied by adding an extra axiom like $\mu(1) = 1$ (so we are working with probability measures in a sense), but it seems highly doubtful even in that case that a left adjoint out of $Bool$ would exist. As noted above, it could only exist if $MBool$ lacked many limits like uncountable products, and this would rule out a straightforward application of an adjoint functor theorem.

I am not sure about a left adjoint out of $Met$.

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Thanks. Very useful insight. –  Davide Bernardini Oct 18 '12 at 14:32

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