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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.
5
votes
Accepted
Countably compact Boolean algebras versus distributivity
There are many countably distributive complete Boolean algebras, and this is an important concept in forcing. For example, the canonical forcing to add a Cohen subset (or any number of Cohen subsets) …
- 236k
5
votes
Accepted
Projections between complete boolean algebras
The answer is no. Let $P$ arise from product forcing $Q\times Q$. So forcing with $P$ adds two mutually generic filters for $Q$, one on each factor. Let $\sigma$ be the projection onto the first coord …
- 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
6
votes
When are two forcing posets "the same"?
Let me augment Calliope's excellent answer by providing a slightly stronger example. What I want to provide is an example exhibiting property (iii) without property (i), but in a strong way, in that t …
- 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
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
1
vote
Density and compactness of Boolean embeddings
Regarding the dense embedding, perhaps this is helpful. Statement 1 can be taken as a definition of density, which makes the connection with topology by means of the lower-cone topology.
Theorem. Supp …
- 236k
4
votes
Accepted
Is the category of atomless Boolean algebras with complete embeddings closed under coproducts?
This category does not have co-products. To see this, let
$\newcommand\B{\mathbb{B}}\B$ be any atomless complete Boolean algebra with a nontrivial automorphism $\pi:\B\to\B$. For example, the forcing …
- 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
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
3
votes
What's "serialization" really called, and is there any theory surrounding it?
Your concept is similar to (but not exactly the same as) the concept of diagonal union, defined for ordinals $\kappa$ and sets $A_\alpha\subset\kappa$ for $\alpha<\kappa$ as follows:
$$\mathop{\bigtr …
- 236k
5
votes
Weak equivalence over forcing notions
Let me provide the alternative kind of example.
Theorem. If ZFC is consistent, then there is a model of ZFC in which any two weakly equivalent forcing notions are forcing equivalent.
Proof. Consid …
- 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
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
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