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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.
27
votes
Accepted
Jonsson Boolean algebras?
Boolean algebras are never Jonsson.
Suppose that $\mathbb{B}$ is a Boolean algebra of size $\omega_1$. Let $a$ be any element such that neither $a$ nor $\neg a$ is an atom. Note that every element $ …
- 236k
17
votes
Accepted
Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})...
The answer is no.
In the Boolean algebra $P(\omega)/\text{Fin}$ every strictly descending sequence $A_0>A_1>A_2>\cdots$ has a nonzero lower bound, by the famous construction of Hausdorff. Namely, one …
- 236k
12
votes
Accepted
On $V$-decisive and weakly homogeneous forcings
$\newcommand\B{\mathbb{B}}$
Update. The answer is no to all three questions.
Theorem. The following are equivalent for any complete Boolean
algebra $\B$.
$\B$ is $V$-decisive.
For any two conditio …
- 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
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
11
votes
Accepted
Are no infinite subsets of the set of all propositional atoms definable in this structure, e...
It's a nice question. This Boolean algebra, known as the Lindenbaum algebra, is a countable atomless Boolean algebra — it is atomless because we can always take the conjunction of any formula with a n …
- 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
9
votes
Accepted
Does $\aleph_0$-density of regular open algebra entail existence of countable basis?
The answer is no, not necessarily.
For a counterexample, consider the Sorgenfrey line, which is the topology on $\mathbb{R}$ with basis consisting of the half-open intervals $[a,b)$. These are each …
- 236k
8
votes
Accepted
Introducing meets while preserving directed closure
$\newcommand\P{\mathbb{P}}$The answer is no. For a counterexample, consider the following
partial order $\P$. On the bottom layer, we have countably many
incompatible atoms $a_n$ for $n<\omega$. On a …
- 236k
8
votes
Examples for "nice" Boolean algebras that are not complete or not atomic
There is up to isomorphism a unique countably infinite atomless Boolean algebra (by a back-and-forth argument), making this algebra highly canonical. But it cannot be complete, since every infinite B …
- 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
7
votes
Accepted
Cohen algebra (generalization)
Regarding question $1$, it seems that you want to know whether
you've got the unique complete c.c.c. Boolean algebra with density
$\kappa$. The answer is no.
On the one hand, the forcing notion $\tex …
- 236k
7
votes
Generalizations of Boolean posets/lattices
The generalization you seek exists when k itself is a power of 2 (but gives no additional examples). This is because, as Q Yuan points out, the important properties of 2 that you seem to require are …
- 236k
7
votes
Accepted
Is ${\cal P}(\omega)/\text{(fin)}$ a fractal poset?
Yes, $P(\omega)/\text{fin}$ is fractal. If $A\subseteq^* B$ but not equivalent, then the interval $[A,B]$ in $P(\omega)/\text{fin}$ consists of the sets that almost contain $A$ and are almost containe …
- 236k
7
votes
Complete Boolean algebra not isomorphic to a $\sigma$-algebra
Let me elaborate on Simon Henry's nice answer.
What we prove is that the Cohen algebra, the completion of the unique countable atomless Boolean algebra, is not isomorphic to any $\sigma$-algebra. Sin …
- 236k