Timeline for Are there $2^{\aleph_0}$ pairwise non-isomorphic Boolean algebra structures on $\omega$?
Current License: CC BY-SA 4.0
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| Sep 6, 2019 at 19:27 | history | edited | Goldstern | CC BY-SA 4.0 |
added a proof sketch
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| Sep 6, 2019 at 10:31 | comment | added | Robert Furber | @YCor That statement is Theorem 1.3 (for all infinite cardinals including $\kappa = \aleph_0$). Immediately after, Monk also proves that for uncountable $\kappa$, there are $2^\kappa$ isomorphism classes of superatomic Boolean algebras as well, unlike the case of $\aleph_0$ (if the continuum hypothesis fails). | |
| Sep 6, 2019 at 10:27 | comment | added | Robert Furber | @Goldstern Sorry, I deleted my comment almost immediately after posting it because it duplicated what YCor said right under the question, but evidently you replied quicker than I expected. | |
| Sep 5, 2019 at 21:06 | comment | added | YCor | (For my last question I especially have $\kappa=\aleph_0$ in mind, as in the OP's question.) | |
| Sep 5, 2019 at 19:29 | comment | added | Emil Jeřábek | Every unstable countable theory has $2^\kappa$ nonisomorphic models of cardinality $\kappa$ for all uncountable $\kappa$. | |
| Sep 5, 2019 at 19:03 | comment | added | YCor | OK thanks. The last statement (that there are $2^\kappa$ total orderings for which the interval algebras (with genuine intervals) are pairwise non-isomorphic) is proved there? | |
| Sep 5, 2019 at 18:39 | comment | added | Goldstern | @RobertFurber Yes, and there are also several other classes where you need $\kappa$ to satisfy some mild assumptions in order to get $2^\kappa$ many BAs, such as $\kappa>\aleph_0$ or $\kappa=\kappa^{\aleph_0}$. | |
| Sep 5, 2019 at 18:34 | comment | added | Goldstern | By "intervals" I mean actual intervals (more precisely: of the form $[a,b)$ or $[a,\infty):=\{x:a\le x\}$), not Dedekind cuts or convex sets. There are only countably many intervals, and countably many finite unions of intervals, if the base set is countable. | |
| Sep 5, 2019 at 18:23 | comment | added | YCor | But if $L$ is countable then $\mathrm{Int}(L)$ can fail to be countable. So I don't see how this yields $c$ countable Boolean algebras, unless you restrict to intervals that have bounds in $L$. | |
| Sep 5, 2019 at 18:15 | history | answered | Goldstern | CC BY-SA 4.0 |