Timeline for Are there $2^{\aleph_0}$ pairwise non-isomorphic Boolean algebra structures on $\omega$?
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| Feb 24, 2019 at 12:07 | comment | added | YCor | By the way this easy argument works for $k$ arbitrary. Let me expand in BA language. (1) Easy lemma: Let $X$ be a set and $A,B$ Boolean subalgebras of $2^X$ both containing the ideal of finite subsets. Then any isomorphism $A\to B$ extends to a automorphism of $2^X$ (thus induced by a permutation of $X$). (2) For a nonprincipal ultrafilter $u$ on $X$ let $M_u$ be the corresponding maximal ideal. For two $u,v$, let $N_{u,v}$ be the fibre product, namely $\{0,1\}+(M_u\cap M_v)$. Then it follows that $N_{u,v}\simeq N_{u',v'}$ iff $(u,v)$ is in the orbit of $(u',v')$ modulo a permutation of $X$. | |
| Feb 23, 2019 at 18:02 | comment | added | YCor | Freniche says, about the case $k=\aleph_0$ that it's "almost trivial and was independently noted by E. K. van Douwen (partial results were obtained independently by S. Williams)." I couldn't track if van Douwen's result's been written. Anyway, here's an argument: first, the homeomorphism group $G$ of $\beta\mathbf{N}$ has cardinal $c$. For any pair $\{x,y\}\subset\beta\mathbf{N}-\mathbf{N}$, let $K_{x,y}$ be the space obtained from $\beta\mathbf{N}$ by gluing $x$ and $y$. It's easy to check that it retains $\{x,y\}$ modulo $G$. Hence this yields $2^c$ non-homeomorphic separable Stone spaces. | |
| Feb 23, 2019 at 17:49 | comment | added | YCor | The relation between the algebra and $\omega$ is the equality relation! it meant an algebra whose underlying set is equal to $\omega$. | |
| Feb 23, 2019 at 17:00 | comment | added | YCor | I see. This terminology is dire, but I agree it exists. Hopefully the OP will soon clarify his request. | |
| Feb 23, 2019 at 16:57 | comment | added | juan | One speaks of algebras, $\sigma$-algebras of parts of a set, as algebras and $\sigma$-algebras in that set. What are your meaning? the elements of the algebra what relation have to $\omega$ in your sense? | |
| Feb 23, 2019 at 16:42 | comment | added | YCor | This does not answer the question. It's a question about the number of Boolean algebra structures on $\omega$, not the number of Boolean subalgebras of $2^\omega$ up to isomorphism. In particular, $2^{\aleph_0}$ is a trivial upper bound. | |
| Feb 23, 2019 at 16:39 | history | edited | juan | CC BY-SA 4.0 |
added 9 characters in body
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| Feb 23, 2019 at 16:30 | history | edited | juan | CC BY-SA 4.0 |
latex correction
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| Feb 23, 2019 at 16:30 | history | edited | Noah Schweber | CC BY-SA 4.0 |
edited body
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| Feb 23, 2019 at 16:27 | history | answered | juan | CC BY-SA 4.0 |