0
votes
0answers
14 views

Boolean Algebras and Spaces

Show that a countably infinite free Boolean algebra $B$ has a Boolean space homeomorphic to $2^\omega$; where $2$ is the discrete space $\{0,1\}$; hence B is isomorphic to the Boolean algebra of ...
0
votes
1answer
22 views

A boolean algebra is complete iff its Stone Space is extremally disconnected

I have the following proof, but I don't understand one of the steps: Theorem 4.4. A Boolean algebra is complete iff its Stone space is exlremally disconnected. Proof. Identify the given ...
0
votes
0answers
15 views

A filterbase generating filter F

Show that a non empty subset $X$ of a filter $F$ in $B$ is a base for $F$ iff $X$ generates $F$ and for all $x,y$ $\in$ $X$ $\exists$ $z $ $\in$ $X$ such that $z$ $\leqq$ x $\wedge$ y.
1
vote
1answer
22 views

Injectivity of Stone embedding

Let $\mathcal{B}$ be a Boolean algebra, $X$ the set of ultrafilters of $\mathcal{B}$ and $\sigma:\mathcal{B}\longrightarrow\mathcal{P}(X)$ the map sending $b\in\mathcal{B}$ to the set of ultrafilters ...
1
vote
1answer
29 views

If $G$ is a generic ultrafilter, why $(\exists a\in A)(a\in G)\leftrightarrow \Sigma A\in G$?

Let $B$ be a complete Boolean Algebra. Let $G$ be a generic ultrafilter of $B$, that is, such that for any dense $D\subset B$ we have $D\cap G\neq \emptyset$. Why for all $A\subseteq B,$ $\Sigma A\in ...
2
votes
1answer
60 views

Filters of Boolean algebras which are Boolean algebras

Looking at some filters generated by elements of a finite Boolean algebra I have the impression that many/most/all of them are by themselves Boolean algebras (at least I didn't stumble upon a ...
0
votes
1answer
156 views

Three questions on chapter 7 of Jech's Set Theory

In the proof of Pospisil's Theorem (theorem 7.6) that there are $2^{2^\kappa}$ uniform ultrafilters on $\kappa \geq \omega$, the author writes : Let $\mathcal{A}$ be an independent family of subsets ...
3
votes
1answer
136 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 $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...