2
votes
0answers
94 views

Back-and-Forth Argument vs. “One-Way” Argument

The wikipedia article on the Back and Forth Argument claims at the end: If we iterated only step $(1)$, rather than going back and forth, then in some cases the resulting function from A to B ...
2
votes
1answer
60 views

Can the obvious “product” of complete atomistic Boolean algebras be realized as a categorial product?

Let $X$ and $Y$ denote sets, and $\eta_X,\eta_Y : X,Y \rightarrow X+Y$ denote the natural injections to the disjoint union. Then intuitively, the "product" of the Boolean algebras $2^X$ and $2^Y$ ...
0
votes
0answers
40 views

Definition of a linear extension (total order?) of a poset

Hey I have a question about the definition of a linear extension of a poset. If I was given a hasse diagram of a poset with relation <= (S, <=), how can I get the compatible total order of this ...
1
vote
1answer
91 views

Atomic Boolean lattice is weakly atomic

The book "Introduction to Lattices and Order" says at Exercise 10.12 that an atomic Boolean lattice is weakly atomic. Could you tell me why it holds?
1
vote
1answer
98 views

The cardinality of finite Boolean lattice

A book (Introduction to Lattices and Order) says that we can prove that the cardinality of any finite Boolean lattice $B$ is a power of 2 by induction on $|B|$ with the following lemma: If a lattice ...
3
votes
1answer
148 views

Do filters on a Boolean algebra also make a Boolean algebra?

Let $\mathfrak{B}=(B,\bot,\top,\lnot,\wedge,\vee)$ be a boolean algebra. $B_F$ be the set of all filters on $\mathfrak B$. And for all filter $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...
3
votes
1answer
306 views

Cardinality of the set of ultrafilters on an infinite Boolean algebra

Let $\mathfrak B$ be a Boolean algebra with an infinite power $\kappa$. My question is how many ultrafilters does it have? $\kappa$ or $2^\kappa$? Or even smaller?
3
votes
1answer
134 views

Inducing maps between Boolean completions of posets

Given a separative poset $P$, we can form it's Boolean completion $B(P)$. This is a Boolean algebra whose elements are regular cuts on $P$, as defined here. Also, $P$ embeds densely into $B(P)$ by ...
4
votes
2answers
161 views

Stone duality for ideals and filters (exercise)

In A Course in Universal Algebra (Burris, Sankapannavar), the exercise 4.4.7-8, p.158, says: Let $A$ be a Boolean algebra. Denote $A^\ast:=\{\text{ultrafilters of }A\}$, and give $A^\ast$ the ...