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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.

16 votes
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Is Stone-Čech compactification of 0-dimensional space also 0-dimensional?

In this paper (Spaces $N\cup\mathscr{R}$ and their dimensions, Topol. Appl. 11(1) 1980 93-102) — I hope the PDF is freely available) Jun Terasawa constructs maximal almost disjoint families on $\mathb …
KP Hart's user avatar
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9 votes
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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 spac …
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7 votes
Accepted

Can any poset of cardinality $\leq 2^{\aleph_0}$ be embedded in ${\cal P}(\omega)/(\text{fin...

Here is an attempt at a 'definitive summary'. To begin with positive results: $\mathsf{CH}$ implies a “yes” answer to this question. The fastest way to see this is to first embed a given partial order …
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6 votes

A unique ultrafilter extending a union of filters?

The following is due to Alan Dow: In any model obtained by adding $\aleph_2$ many Cohen reals to a model of $\mathsf{CH}$ the statement is false. We force with $\mathbb{P}=\operatorname{Fn}(\omega_2, …
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6 votes
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Can the Boolean Algebra of regular open sets be isomorphic to ${\cal P}(\omega)/(\text{fin})$?

No, $\mathrm{RO}(X)$ is complete; $\mathcal{P}(\omega)/\mathit{fin}$ is not (no strictly increasing sequence has a supremum).
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5 votes

Self-homeomorphism of Stone-Čech boundary with an isolated fixed point

As an answer to the bonus question: no, see K. P. Hart and J. Vermeer. Fixed-point sets of autohomeomorphisms of compact F-spaces, Proceedings of the American Mathematical Society, 123 (1995), 311–314 …
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5 votes
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A problem of non-emptiness of intersections of certain chains of regular open sets

Here is a provisional negative answer. If $\mathcal{C}$ is a c-minimal chain then $N=\bigcap\mathcal{C}$ is closed and nowhere dense (if nonempty): if the interior is nonempty take a nonempty regular …
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5 votes
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Stone-Čech compactification of a Boolean subalgebra of $\{0,1\}^S$

A partial answer. for your specific example, the algebra $B$ generated by the rational intervals, the answer is negative. This algebra is countable and dense in $2^\mathbb{R}$. It is countable and ze …
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4 votes

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

This question was answered in topological terms by J. Vermeer in The smallest basically disconnected preimage of a space. Topology Appl. 17 (1984), no. 3, 217–232. See here for a review and here for t …
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4 votes
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Extremally disconnected rigid infinite Hausdorff compacta(?)

This paper, Rigid Stone spaces within $\mathsf{ZFC}$ by Dow, Gubbi, and Szymański contains relatively easy examples of such spaces.
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4 votes
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Copy of $P(\omega)/\mathrm{fin}$ on $\omega_1$

To expand my comment into an answer: take, for each $n\in\omega$, a uniform ultrafilter $u_n$ on $\omega_1$ that contains the set $\{\lambda+n:\lambda$ is a limit or $0\}$. The set $U=\{u_n:n\in\omega …
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3 votes

Chain of ideals in a BA

Take te interval algebra, $B$, of the ordinal $\omega_1\times\omega$; for $n\in\omega$ let $I_n$ be the ideal consisting of the elements of $B$ that are bounded below $\omega_1\times n$. None of the i …
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3 votes

Ideals on $\mathbb N$ and large sets that have small intersection

In topological language: for any closed subset, $F$, of $\beta\mathbb{N}\setminus\mathbb{N}$ the family $I_F=\{A:A^*\cap F=\emptyset\}$ is an ideal. Your property translates into: the closed set $F$ i …
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2 votes

Can this ultrafilter convergence condition be expressed as a compactness condition?

Your condition amounts to saying that the uniformity $\mathcal{C}$ generated by all clopen partitions is complete: 'partition-prime' is the same as '$\mathcal{C}$-Cauchy'. It is also implied by the co …
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2 votes
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Description of atomless complete Boolean algebras with a countable $\pi$-base

To make the answers concrete: the Boolean algebra of clopen subsets of the Cantor set is the unique countable atomless Boolean algebra. Its completion is the regular-open algebra of the Cantor set, wh …
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