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Results tagged with boolean-algebras
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user 61785
A Boolean algebra is a commutative ring satisfying x²=x for every x, and sometimes required to have a unit; they have characteristic 2. For coding theory (notably dealing with subsets linear subspaces of spaces of Boolean functions), rather use the [coding-theory] or [linear-algebra] tag.
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Representation of free Boolean sigma-algebras
There appears to be some confusion in the question, which I will try to dispel.
As a free Boolean $\sigma$-algebra $A$ is a Boolean $\sigma$-algebra, the Loomis-Sikorski theorem certainly applies. Ho …
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Finitely additive measures on Boolean algebras of regular open subsets: Is there a relations...
The fact you are probably looking for is that, for any Baire space $X$ (e.g. a completely metrizable space or a compact Hausdorff space) the inclusion map $\mathfrak{R}(X) \rightarrow \mathfrak{B}o(X) …
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What is to Stone space of the free sigma-algebra on countably many generators?
You got a wrong answer on Math Stackexchange from Daron. The free Boolean algebra on countably many generators is the Boolean algebra of clopens of $2^\omega$ (topologized with the product topology), …
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Is the boundary of an open set in a $\sigma$-space empty?
In effect, you are asking if $\bigvee\limits_{i \in \omega}B_i = \bigcup\limits_{i \in \omega}B_i$, where the left hand side is the closure of the union, which is the join/supremum of the family $\{B_ …
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Infinite distributive laws in atomless free sigma-algebra
This holds because $\mathfrak{A}$ is a concrete $\sigma$-algebra, being the Baire $\sigma$-algebra of $2^{\omega_1}$. In fact, the cardinality of $\omega_1$ plays no role whatsoever and $\omega_1$ cou …
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Functor from rings into compact Hausdorff spaces
Here is an answer to your question about monadicity, as it's too long for a comment. I will not fill in every detail, so if you follow along there will be several definitions that need to be expanded …
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Is a Boolean algebra with an order continuous topology a measure algebra?
It is not true that $B$ is necessarily a measure algebra. The counterexample is due to Michel Talagrand, who constructed a Maharam algebra that is not a measure algebra.
Maharam, D., An algebraic cha …
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