4
$\begingroup$

Given a family of Boolean algebras $\mathcal{B}=\{B_i\colon i\in I\}$ with respective Stone spaces $S_i$, the algebra of clopen (both closed and open) subsets of the product space $\textstyle\prod_{i\in I}S_i$ is called the free product of $\mathcal B$. This algebra is typically denoted by $\textstyle\bigotimes_{i\in I}B_i$ (and I will use the standard "tensor" notation for finite free products in the obvious manner).

I am interested in the (possible) Boolean algebras which admit only very particular decompositions in terms of the free product.

Is there an uncountable Boolean algebra $B$ such that if $B$ is isomorphic to $A\otimes C$ then either $A$ or $C$ is countable?

$\endgroup$
2
  • 2
    $\begingroup$ Remark: This tensor product is just the usual tensor product of algebras over $\mathbb{F}_2$. $\endgroup$ – Martin Brandenburg Jul 21 '11 at 10:26
  • $\begingroup$ Wow, a downvote for a question from 2011! $\endgroup$ – Tomasz Kania Mar 21 '19 at 19:58
10
$\begingroup$

Translating this to Boolean spaces, you are looking for a Boolean space X which is not second countable, but cannot be written as a product of two factors of the same type (i.e., not second countable).

Have you considered the compact space $[0,\omega_1]$? It is certainly not the product of two uncountable spaces, as such a product would contain two almost disjoint closed uncountable sets. On the other hand, a countable Boolean space cannot have uncountably many clopen sets.

$\endgroup$
1
  • 1
    $\begingroup$ It's not even a product of two spaces of cardinal $\ge 2$. The reason is that every point, with a single exception, has a countable neighborhood. $\endgroup$ – YCor Mar 21 '19 at 9:32
5
$\begingroup$

By Theorem 15.14 of the Boolean algebra handbook, the interval algebra of the real numbers is such an algebra B.

Reference: S. Koppelberg. Handbook of Boolean algebras. Vol. 1. Edited by J. D. Monk and R. Bonnet. North-Holland Publishing Co., Amsterdam, 1989.

$\endgroup$
2
  • 2
    $\begingroup$ For completeness, $B$ here is the Boolean subalgebra of subsets of $\mathbf{R}$ generated by the left-closed, right-open intervals. Its Stone space can be viewed as the doubled circle $S^{\pm}$, starting from $\mathbf{R}\cup\{\infty\}$, with each point $x$ replaced by two points $\{x^-,x^+\}$, with the topology of circular order. Theorem 15.14 in the handbook even shows that $B$ doesn't contain the free product of an infinite subalgebra and an uncountable one. In other words, there is no continuous surjection from $S^{\pm}$ onto the product of an infinite with a non-metrizable Stone space. $\endgroup$ – YCor Mar 21 '19 at 8:15
  • 2
    $\begingroup$ Also note that unlike Goldstern's example (whose space is not a product of any two spaces of cardinal $\ge 2$), this space is homeomorphic to its product with any nonempty discrete finite set. $\endgroup$ – YCor Mar 21 '19 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.