# Tagged Questions

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### 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 ...
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### 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 $2^Y$ ...
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### 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 ...
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### Can a brouwerian lattice be extended into a boolean algebra?

Can an arbitrary brouwerian lattice (=locale = frame) be extended into a boolean algebra? What do I mean by "extended"? I don't know. All I know is that our brouwerian lattice is a sub-poset of the ...
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### 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?
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### 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 ...
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### 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 $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...
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### 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?
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 ...
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 ...