Timeline for Complete Boolean algebra not isomorphic to a $\sigma$-algebra
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2018 at 12:36 | history | edited | Simon Henry | CC BY-SA 3.0 |
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| Jan 15, 2018 at 10:30 | comment | added | Simon Henry | @IoannisSouldatos : I agree with you. the reason I didn't really explain it is because it is a "well known fact" in some area which I was familiar with : In operator algebra it corresponds as I said to the fact that von Neuman algebra are "monotone complete" (arbitrary directed subset of bounded positive self-adjoint element have supremum). In topos theory it is closely related to the historical first example of a topos without points. But this is indeed far from being obvious. | |
| Jan 15, 2018 at 9:36 | comment | added | Ioannis Souldatos | @SimonHenry. I think this will work, but at least for me, the fact that the equivalence class of an uncountable union is not the same as the uncountable union of the individual equivalence classes was not obvious. | |
| Jan 10, 2018 at 22:05 | comment | added | Simon Henry | For exemple an uncountable supremum of singleton is the still the empty set. If you want a more explicit proof that it is complete, the point is that if you have an ascending chain indexed by an ordinal, if the ordinal is countable then you just take the union, if the ordinal is uncountable then the measure of each terms is an uncountable increasing sequence of real number smaller than $1$, and such a sequence is constant after some rank, and so the sequence became constant up to the equivalence relation after some rank, so it has a supremum. | |
| Jan 10, 2018 at 22:03 | comment | added | Simon Henry | @IoannisSouldatos : I stand by what I said : $\Sigma$ is isomorphic as an ordered set to the subset of the algebra $L^{\infty}([0,1],\mu)$ of equivalences class of essentially bounded function up to equality almost everywhere. It is the subset of function with value (almost everywhere) in $\{0,1\}$ and this algebra has (as every von Neuman algebra) all supremum and this subset is closed under supremum (because it is characterized by $x^2=x$. The point is that uncountable supremum of element of $\Sigma$ might have nothing to do with union of sets(...) | |
| Jan 10, 2018 at 7:39 | comment | added | Ioannis Souldatos | I know it is too late for this comment, but $\Sigma$ is a countably complete boolean algebra, not complete in general, as this would imply that all subsets of reals are Lebesgue measurable. | |
| Feb 21, 2014 at 21:02 | vote | accept | Bjørn Kjos-Hanssen | ||
| Feb 21, 2014 at 20:01 | comment | added | Simon Henry | No, I do not assume that $X$ is a subset of [0,1], when I said that $x$ belongs to $[a/k,(a+1)/k]$ I meant " x belong to the subset of $X$ which is identified with the object of $\Sigma$ corresponding to $[a/k,(a+1)/k]$ " | |
| Feb 21, 2014 at 19:21 | comment | added | Joel David Hamkins | + 1. Very nice. | |
| Feb 21, 2014 at 19:10 | comment | added | Bjørn Kjos-Hanssen | Can we assume $X\subseteq [0,1]$? | |
| Feb 21, 2014 at 19:02 | history | answered | Simon Henry | CC BY-SA 3.0 |