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27 votes
Accepted

In the category of sigma algebras, are all epimorphisms surjective?

$\require{AMScd}$ A 1974 paper of R. Lagrange, Amalgamation and epimorphisms in $\mathfrak{m}$-complete Boolean algebras (Algebra Universalis 4 (1974), 277–279, DOI link), settled this affirmatively. ...
Badam Baplan's user avatar
20 votes
Accepted

Is every complete Boolean algebra isomorphic to the quotient of a powerset algebra?

This variation of the question comes from the comments on the original question. The question is whether all (complete) BAs are isomorphic in the category BA to a quotient of a powerset algebra. The ...
Keith Kearnes's user avatar
19 votes
Accepted

What is a module over a Boolean ring?

Theorem: Given $A$ a boolean ring/boolean algebra then there is an equivalence of categories between the category of $A$-modules and the category of sheaves of $\mathbb{F}_2$-vector spaces on Spec $A$....
Simon Henry's user avatar
  • 40.2k
18 votes

In what sense is GCD an extension of boolean OR?

In the ordering $\preceq$ of nonnegative integers by divisibility, 1 is the least element and 0 is the greatest, and we have for instance $$ 1\preceq 2\preceq 6\preceq 12\preceq\dots\preceq 0.$$ In ...
Bjørn Kjos-Hanssen's user avatar
17 votes
Accepted

Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic?

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. ...
Joel David Hamkins's user avatar
16 votes

Near permutation $n\mapsto n+1$ not conjugate to its inverse on the Stone-Čech remainder?

Update: The answer is yes -- if $\mathsf{CH}$ is true then $\phi$ and $\phi^{-1}$ are conjugate in the group of self-homeomorphisms of $\omega^*$. I've written this up in a new paper, which you can ...
Will Brian's user avatar
  • 17.5k
15 votes
Accepted

Is it possible to completely embed complete Heyting Algebras into upsets of a poset?

No, not in general: for instance, the real interval $([0,1],{\le})$, or any non-atomic complete Boolean algebra, do not have such an embedding. This follows from the following characterization: ...
Emil Jeřábek's user avatar
14 votes
Accepted

Linear suborders of $(P(\omega),\subseteq)$

For any $a\neq b\in\omega$ pick a set $A_{ab}\in L$ such that $a\in A_{ab},b\not\in A_{ab}$, if such exists, and let $D$ be the set of all $A_{ab}$ we have picked. We claim $D$ is dense in $L$. Let $...
Wojowu's user avatar
  • 27.4k
14 votes
Accepted

Complete Boolean algebras of subsets of $\mathbb N$

The answer is negative. Let $A$ be the completion of the denumerable atomless BA $B$. Then $A$ is complete and atomless. $A$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$. In fact, $B$ can ...
Don Monk's user avatar
  • 156
12 votes
Accepted

Are there functions $\mathbb{F}_2^n \to \mathbb{F}_2$ satisfying these special relations?

I think the answer is no. To see this, observe that \begin{equation*} \left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2=\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\mathbb{E}_{\mathbf{y}}[(...
Jason Gaitonde's user avatar
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 ...
Joel David Hamkins's user avatar
11 votes

Linear suborders of $(P(\omega),\subseteq)$

Note that the lexicographic order of $2^\omega$ is a linear extension of $(\mathcal P(\omega),\subseteq)$, namely if $A\subseteq B$, then $\min(A\mathbin{\triangle}B)\in B$, which means that the first ...
Asaf Karagila's user avatar
  • 38.2k
11 votes
Accepted

Are no infinite subsets of the set of all propositional atoms definable in this structure, even with parameters?

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 ...
Joel David Hamkins's user avatar
11 votes
Accepted

Is every homogeneous poset a lattice?

Counterexample. Let $$P=\{(x,i)\in\mathbb Q\times\{0,1\}:0\le x\le1,\ x\ne i\}$$ be ordered so that $$(x,i)\lt(x',i')\iff x\lt x'.$$
bof's user avatar
  • 11.9k
10 votes
Accepted

Are there $2^{\aleph_0}$ pairwise non-isomorphic Boolean algebra structures on $\omega$?

The answer is yes: there are $2^{\aleph_0}$ countable Boolean algebras up to isomorphism, or equivalently $2^{\aleph_0}$ homeomorphism class of metrizable totally disconnected compact Hausdorff spaces....
YCor's user avatar
  • 60.4k
10 votes

Analytical origins of the Stone duality

In addition to the historical component, the question also asked about the relation between spectral theory and Stone spaces. $\def\Z{{\bf Z}} \def\R{{\bf R}} \def\C{{\bf C}} \def\Spec{\mathop{\rm ...
Dmitri Pavlov's user avatar
10 votes

When are two forcing posets "the same"?

I shall refer to the condition that you describe in Question 1 as condition (a). If you have condition (a) then you get back (ii), since if $B\upharpoonright b\simeq C\upharpoonright c$ then for all $...
Calliope Ryan-Smith's user avatar
9 votes
Accepted

How "strong" is the existence of a non trivial ultrafilter on $\omega$?

It seems difficult to make the notion of "general choice principle" precise, but I would guess that the principle "every infinite set admits a nonprincipal ultrafilter" would qualify. If so, then it ...
Andreas Blass's user avatar
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 ...
Joel David Hamkins's user avatar
9 votes
Accepted

Semi-rigid boolean algebras

There is no such algebra. In fact, suppose that $B$ satisfies the indicated condition. Choose $a$ in $B$ with $a$ not equal to $0$ or $1$. Let $f$ be a nontrivial automorphism of the principal ideal ...
Don Monk's user avatar
  • 883
9 votes

Boolean algebra of ambiguous Borel class

This is a very interesting question whose answer depends on dimension properties of the spaces $X,Y$. First we introduce a suitable terminology. A function $f:X\to Y$ between topological spaces is ...
Taras Banakh's user avatar
9 votes
Accepted

A strictly descending chain of subalgebras of $P(\omega)/_{\mathrm{fin}}$

There is a family $\{K_X:X\subseteq\mathfrak{c}\}$ of separable compact zero-dimensional spaces such that there is a continuous surjection of $K_X$ onto $K_Y$ if and only if $X\subseteq Y$. These ...
KP Hart's user avatar
  • 10.1k
8 votes
Accepted

A curiosity on complete homomorphisms of boolean algebras

An example in which $i$ is a complete homomorphism but $\pi$ is not an open map: Notation: For any Boolean algebra $A$ and any $a$ in $A$, $S(a)$ is the set of all ultrafilters $F$ such that $a$ is in ...
Don Monk's user avatar
  • 883
8 votes
Accepted

Complete atomless Boolean algebras with abelian automorphism group

If A is a rigid complete BA, then the automorphism group of AxA is isomorphic to the direct sum of |A| copies of the two-element group.This is proved in the remark following Lemma 1.9 in a paper of ...
Don Monk's user avatar
  • 883
8 votes

Reduced products of complete Boolean algebras

If you write $2$ for the $2$-element Boolean algebra, and $F$ for the filter of cobounded sets in $\kappa$, then $2^\kappa/F$ is just $P(\kappa)/[\kappa]^{<\kappa}$, which is certainly not complete....
Goldstern's user avatar
  • 14k
8 votes
Accepted

A set theoretic question arising from trying to understand a sheaf cohomology question

This is not onto for any uncountably infinite $T$, even one much smaller than the power set (if the continuum hypothesis is false). Fix $X \in 2^{ S \times T}$ such that the induced map $h \colon T \...
Will Sawin's user avatar
  • 137k
7 votes

Is there such a thing as the sigma-completion of a Boolean algebra?

In the course of a forthcoming research project with Asgar Jamneshan (in ergodic theory), we managed to discover a rather explicit answer to this question, which I am recording here if anyone is ...
Terry Tao's user avatar
  • 109k
7 votes

Does $\aleph_0$-density of regular open algebra entail existence of countable basis?

Joel has already shown you that none of the standard separation axioms is enough to force a positive answer to your question (by the way, the Sorgenfrey line is even perfectly normal). A notable case ...
Santi Spadaro's user avatar
7 votes

Is it possible to completely embed complete Heyting Algebras into upsets of a poset?

Emil Jeřábek gave more complete answer than mine, so I have initially deleted it. On the afterthought, I decided to turn it into an addendum to Emil's answer. It is a generalization of sorts: if an ...
მამუკა ჯიბლაძე's user avatar
7 votes

Analytical origins of the Stone duality

Stone himself gave a brief account of his discovery of Stone duality in a letter written in 1976. Apparently, von Neumann was involved. A reminiscence on the extension of the Weierstrass approximation ...
Dmitri Pavlov's user avatar

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