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Results tagged with boolean-algebras
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user 1946
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.
3
votes
Incomplete subsets of the free boolean algebra on countably many generators
Yes, that claim in the post is wrong, and you are correct to object. What is true — and I was very surprised to learn this — is that in the free Boolean algebra on a countably infinite set of generato …
- 236k
4
votes
Accepted
extending $\sigma$-complete boolean homomorphism
If the Boolean algebra $A$ is c.c.c., then of course every
$\sigma$-complete dense subalgebra $B$ is all of $A$, and so in this case the
answer is trivially affirmative.
If $A$ is not c.c.c., however …
- 236k
5
votes
Quasi-dense subsets of boolean algebras
Well, this still doesn't answer the atomless question, but I've got a violation of the desired implication among atomic Boolean algebras.
Let $A$ consist of the finite or cofinite subsets of $\mathb …
- 236k
10
votes
Is there such a thing as the sigma-completion of a Boolean algebra?
Every Boolean algebra $\mathbb{B}$ embeds densely in its completion as a Boolean algebra, which is a complete Boolean algebra (more than just countably complete). The completion $\bar{\mathbb{B}}$ can …
- 236k
6
votes
Sigma-complete Lindenbaum algebras?
There is no countably infinite $\sigma$-complete Boolean algebra. Once the Boolean algebra is infinite, it must have an countably infinite antichain $A\subset\mathbb{B}$, and then by $\sigma$-complete …
- 236k
7
votes
Accepted
On intermediate transitive models for ZFC between M an M[G]
It depends on the particular forcing, and in general, things may
not work out so nicely.
On the one hand, it could be that $P=\mathbb{B}^+$, in which case
for any intermediate model $N$ we have $X=Y= …
- 236k
11
votes
Accepted
An exercise in Jech's Set Theory
Your counterexample is not correct. Let $r$ be an irrational real number, and let $F$ be the principal ultrafilter in $B$ on the closed interval $[r,r]=\{r\}$, which is an atom in $B$. Note that $F\ca …
- 236k
4
votes
Antichains and measure-preserving actions on Boolean algebras
Regarding the question is your last paragraph, there is the following often-studied but not-quite-equivalent-to-your property:
A Boolean algebra $\mathbb{B}$ is almost homogeneous if for every nonze …
- 236k
2
votes
Chain of ideals in a BA
Let $\mathfrak{A}$ be the power set of an uncountable set $X$, which is a complete Boolean algebra. Select disjoint sets $X_n\subset X$ of size $\omega_1$, and let $J_n$ be the ideal generated by $X_0 …
- 236k
5
votes
Accepted
completions of regular suborders
$\newcommand\P{\mathbb{P}}\newcommand\Q{\mathbb{Q}}
\newcommand\Q{\mathbb{Q}}\newcommand\B{\mathbb{B}}\newcommand\Z{\mathbb{Z}}
\newcommand\RO{\text{RO}}$Unless I am mistaken (and please correct me if …
- 236k
3
votes
Is every c.c.c. non-atomic partial order of size $\omega_1$ a union of countable complete su...
Noah's affirmative answer is correct with the definition that you've given, but if you want to insist that the suborder is also non-atomic, then the answer can be negative.
One easy way to see this …
- 236k
5
votes
Accepted
perfect space without convergent long sequences
It seems that there is no such space. Indeed, I claim that every non-isolated point in a Boolean space is the limit of a long sequence in that space.
We may assume that the space $X$ is the Stone sp …
- 236k
6
votes
subalgebra of a simple forcing
Update. If $\alpha$ is countable, then I claim that $A$ and
$B$ are forcing equivalent, the quotient forcing is atomic,
isomorphic to $P(\omega_1)$, and every extension by $A$ adds a
generic for $\tex …
- 236k
6
votes
Accepted
density of boolean algebras
Yes, it is possible.
Let $\mathbb{B}$ be any complete Boolean algebra with density $\aleph_0$. For example, we could use the power set Boolean algebra $\mathbb{B}=P(\mathbb{N})$, which has density $\ …
- 236k
11
votes
Accepted
Boolean ultrapower of V[G] by G
I share your view that this is a subtle point. To illustrate it, my co-author Dan Seabold and I had pointed to the case of adding a Cohen subset to $\omega_1$ (see example 44 in Boolean ultrapowers pa …
- 236k