Timeline for Ordered union of Borel sets
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Mar 30, 2019 at 5:50 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:50 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:50 | |||||
| Mar 30, 2019 at 5:49 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:50 | |||||
| Mar 30, 2019 at 5:49 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:49 | |||||
| Mar 30, 2019 at 5:49 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:49 | |||||
| Mar 30, 2019 at 5:49 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:49 | |||||
| Mar 28, 2019 at 21:57 | comment | added | Andreas Blass | @Skeeve Whether or not CH holds, every uncountable Borel set (in a separable metric space) includes a homeomorphic copy of the Cantor set and therefore has the cardinality of the continuum. (See for example Theorem 13.6 in Kechris's book "Classical Descriptive Set Theory" or Corollary 2C.3 of Moschovakis's book "Descriptive Set Theory".) This holds whether or not CH holds, but it's useful in my earlier comment just when CH fails. | |
| Mar 28, 2019 at 21:31 | comment | added | Skeeve | @AndreasBlass maybe I am missing something obvious, but if CH fails, why all uncountable Borel sets have the cardinality of continuum? It seems to me the opposite: if CH holds then the cardinality of any uncountable Borel set is at most $2^{\aleph_0} = \aleph_1$ and at least $\aleph_1$. | |
| Mar 28, 2019 at 17:04 | comment | added | YCor | An algebraic synthesis of some already given answers: it is a simple fact that a Boolean subalgebra of a given complete Boolean algebra stable under taking arbitrary supremums, if and only if it is stable under taking supremums of (well-ordered) chains. Hence, in a $T_0$ topological space ($T_0$ means that singletons are Borel), the Borel Boolean subalgebra has the property of being stable under taking unions of chains iff it is the whole power set. | |
| Mar 28, 2019 at 15:58 | answer | added | Burak | timeline score: 2 | |
| Mar 28, 2019 at 13:38 | history | edited | YCor | CC BY-SA 4.0 |
added 3 characters in body; edited title
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| Mar 28, 2019 at 12:53 | comment | added | Nate Eldredge | @AndreasBlass: Indeed, if the continuum hypothesis holds, you may simply take the Vitali set or your other favorite non-Borel set; it will automatically have cardinality $\aleph_1$. | |
| Mar 28, 2019 at 12:29 | comment | added | Andreas Blass | To complete the answer given by @FrançoisG.Dorais in a comment, note that there is (assuming ZFC) a non-Borel set of size $\aleph_1$. Take a set $X$ of reals of size $\aleph_1$: If the continuum hypothesis holds, then $X$ has more than continuum many subsets of size $\aleph_1$ but there are only continuum many Borel sets. If the continuum hypothesis fails, then $X$ itself isn't Borel because all uncountable Borel sets have the cardinality of the continuum. | |
| Mar 28, 2019 at 11:44 | comment | added | Andrés E. Caicedo | @31415926 For $[0,1]$ that would be equivalent to assuming the continuum hypothesis. | |
| Mar 28, 2019 at 9:53 | answer | added | Skeeve | timeline score: 7 | |
| Mar 28, 2019 at 6:21 | comment | added | user137602 | I am still confused... Say, we have a set $[0,1]$. How do we write it as the union of ordered set $\{A_i:i\in I\}$, in which each $A_i$ is countable? Can you give me an example of these $A_i$? | |
| Mar 28, 2019 at 5:58 | comment | added | James E Hanson | Since the set has size $\aleph_1$, you can write it as a union of a linearly ordered collection of countable sets. All countable sets are automatically Borel. | |
| Mar 28, 2019 at 5:20 | review | Close votes | |||
| Mar 28, 2019 at 18:35 | |||||
| Mar 28, 2019 at 4:58 | comment | added | user137602 | Thank you for the response! Is it possible to provide a reference or proof for your answer? | |
| Mar 28, 2019 at 3:41 | history | edited | user137602 | CC BY-SA 4.0 |
edited title
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| Mar 28, 2019 at 3:40 | comment | added | François G. Dorais | Every set of size $\aleph_1$ can be written in this way, so "no". | |
| Mar 28, 2019 at 3:40 | review | First posts | |||
| Mar 28, 2019 at 6:18 | |||||
| Mar 28, 2019 at 3:38 | history | asked | user137602 | CC BY-SA 4.0 |