Timeline for Is there an uncountable Borel almost disjoint family?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2016 at 20:47 | vote | accept | Richard Rast | ||
| Jan 15, 2016 at 20:00 | answer | added | Goldstern | timeline score: 3 | |
| Jan 15, 2016 at 17:20 | answer | added | Ali Enayat | timeline score: 16 | |
| Jan 15, 2016 at 10:01 | comment | added | Ali Enayat | @RichardRast. Here is an elaboration of Asaf Karagila's first comment: "tag" each vertex of the binary tree $2^{<\omega}$ by a natural number. Then each branch of the tree gives rise to a unique subset of $\omega$, and any two such subsets are clearly almost disjoint. This family is easily seen to be a Borel uncountable almost disjoint family. | |
| Jan 15, 2016 at 6:54 | comment | added | Noah Schweber | @AndrésCaicedo I presume the construction is the Zorn's Lemma one: first show that there is a maximal almost disjoint family via Zorn, and then show that no countable almost disjoint family is maximal almost disjoint. (And in fact there is no Borel, or even analytic, maximal almost disjoint set - this was proved by Mathias in "Happy Families," see the first page of math.uni-hamburg.de/home/khomskii/papers/…) | |
| Jan 15, 2016 at 4:49 | review | Close votes | |||
| Jan 15, 2016 at 10:30 | |||||
| Jan 15, 2016 at 4:32 | comment | added | Andrés E. Caicedo | Which constructions you know of do not result on Borel sets? The standard examples do. | |
| Jan 14, 2016 at 23:23 | comment | added | Richard Rast | Fair point, thank you. Care to turn this into an answer, so it can be accepted? | |
| Jan 14, 2016 at 22:37 | comment | added | Asaf Karagila♦ | Richard, that depends on her w you represent the reals. If they are actual sets of integers, that's one thing. If they are branches through a tree that's another. Branches in $2^{<\omega}$ are sets of finite functions which have some additional property to them, and you can easily show they form an almost disjoint family. And it's not terribly difficult to code this tree into the integers. | |
| Jan 14, 2016 at 19:34 | comment | added | Richard Rast | Asaf, I'm a bit confused about your comment. It seems to me that every real is a branch through that tree, but of course you can have two reals which differ at only a single coordinate. | |
| Jan 14, 2016 at 19:31 | history | edited | Richard Rast | CC BY-SA 3.0 |
Fixed definition of almost disjoint
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| Jan 14, 2016 at 19:30 | comment | added | Richard Rast | Ahhhhhh typos. No idea why I said symmetric difference. Question is fixed. But thank you. Care to make something an answer? | |
| Jan 14, 2016 at 18:29 | comment | added | Andreas Blass | Asaf's example is Noah's example when the language is a propositional language with countably many propositional atoms. | |
| Jan 14, 2016 at 16:03 | comment | added | Noah Schweber | @GeraldEdgar That's easy, though. Being a complete theory (coded appropriately as a subset of $\omega$) is a $\Pi^0_1$ property - in fact, the set of codes for complete theories is closed! | |
| Jan 14, 2016 at 15:56 | comment | added | Gerald Edgar | @NoahSchweber ... then you have to show complete theories are Borel | |
| Jan 14, 2016 at 15:50 | comment | added | François G. Dorais | I think it's a typo: the proposed "weak" definition of almost disjointness leads to families that are not at all disjoint in any typical sense. For example, $\{(-\infty,x)\cap\mathbb{Q}: x \in \mathbb{R}\}$. | |
| Jan 14, 2016 at 15:45 | comment | added | Noah Schweber | Note that using your weaker definition of almost disjoint, you could also just take the set of complete theories in your favorite (nontrivial) language. | |
| Jan 14, 2016 at 15:38 | comment | added | Paul McKenney | Your definition of "almost disjoint" is much weaker than the usual one, which would be "for every pair $X\neq Y$ from $\mathcal{F}$, $X\cap Y$ is finite". In any case, Asaf's example satisfies both definitions. | |
| Jan 14, 2016 at 15:27 | comment | added | Asaf Karagila♦ | Seems relevant: math.uni-hamburg.de/home/khomskii/papers/… | |
| Jan 14, 2016 at 15:23 | comment | added | Asaf Karagila♦ | If you look at the tree $2^{<\omega}$, the set of branches is almost disjoint. | |
| Jan 14, 2016 at 15:16 | history | asked | Richard Rast | CC BY-SA 3.0 |