Timeline for Sigma algebra without atoms ?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2010 at 0:54 | comment | added | Joel David Hamkins | I agree with that, and I don't have any issue with $2^A$. My point rather was that by cutting down one can construct an atomless $\sigma$-algebra living on the reals themselves, rather than living on $2^R$, and this I expect the topologists would like even more. | |
| Apr 27, 2010 at 17:38 | comment | added | Gerald Edgar | Perhaps only a logician would think $\omega_1$ is simpler than the reals. In the topological space $\{0,1\}^A$, no point is extraneous: if you maliciously remove one, any topologist would just put it back in! | |
| Apr 26, 2010 at 12:31 | comment | added | Joel David Hamkins | ... If you use $A=\omega_1$ and then restrict the product space to periodic sequences, then you can still distinguish any two sets in your algebra, but now the underlying set of the algebra has size only continuum, and so you can find an isomorphic copy of the algebra living on the reals. | |
| Apr 26, 2010 at 12:29 | comment | added | Joel David Hamkins | Gerald, I don't understand your remark about going backwards. After all, you can always make the underlying set large by adding extra irrelevant points (e.g. replacing each point with many copies); my question was what is the smallest set of points that would support your $\sigma$-algebra? If you take A to be R, then your space has size $2^{2^\omega}$, which is definitely larger than continuum. But I have a feeling that Cosmonaut or others would prefer an example whose underlying set is a set of reals. | |
| Apr 26, 2010 at 11:57 | comment | added | Gerald Edgar | @Joel: I think cutting down is going backwards. Plus putting structure on the set $A$ is making the example more complicated. If anything, $A$ should be $\mathbb{R}$ and not $\omega_1$. | |
| Apr 26, 2010 at 1:22 | comment | added | François G. Dorais | Gerald's example does a little more: $\mathcal{F}$ is the free $\sigma$-algebra with $|A|$ generators. | |
| Apr 25, 2010 at 22:25 | comment | added | Joel David Hamkins | I suppose that your space is rather large, with size $2^{\omega_1}$ (which I admit may have size only continuum, but often is larger), but can you cut it down by reducing to a "dense" set of size continuum? Perhaps it works if you restrict $2^A$ just to the periodic sequences? | |
| Apr 25, 2010 at 20:53 | comment | added | Joel David Hamkins | Very nice, Gerald. | |
| Apr 25, 2010 at 18:04 | history | answered | Gerald Edgar | CC BY-SA 2.5 |