Timeline for Is the Hausdorff metric on sub-$\sigma$-fields separable?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2020 at 13:36 | comment | added | Dave L Renfro | Maybe someone can fix this? Apparently the substitution principle doesn't hold for URL exchanges. | |
| Jan 16, 2020 at 13:27 | history | edited | Dave L Renfro | CC BY-SA 4.0 |
Original URL to a copy of my sci.math post led to a blocked site when I tried it a few minutes ago, so I've replaced it with a safe URL.
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| Nov 3, 2011 at 14:18 | comment | added | Jason Rute | @Nate Eldredge, I think it is size continuum as follows: If $\rho(\mathcal{G},\mathcal{H})\neq 0$, then $f \mapsto E[f \mid \mathcal{G}]$ and $f \mapsto E[f \mid \mathcal{H}]$ are different operators. But in $L^2$ these operators are continuous linear transformations of which there are only continuum many (correct?). | |
| Nov 3, 2011 at 14:12 | comment | added | Jason Rute | @Gerald Edgar, thanks I thought it might be something like that. | |
| Nov 3, 2011 at 13:51 | vote | accept | Jason Rute | ||
| Nov 3, 2011 at 13:33 | answer | added | Bill Johnson | timeline score: 5 | |
| Nov 3, 2011 at 4:52 | comment | added | Nate Eldredge | As a first step, what is the cardinality of the set of complete $\sigma$-algebras? If it isn't $2^{\aleph_0}$, then that's certainly an obstruction. | |
| Nov 3, 2011 at 4:51 | answer | added | Yuri Bakhtin | timeline score: 0 | |
| Nov 3, 2011 at 0:25 | comment | added | Gerald Edgar | It is called the Hausdorff pseudometric because it is an instance of Haudsorff's construction starting with pseudometric $\mu(A \triangle B)$. In general, Hausdorff's construction starts with a pseudometric and constructs a new pseucometric on subsets of the original space. See here en.wikipedia.org/wiki/Hausdorff_distance . | |
| Nov 2, 2011 at 21:58 | history | asked | Jason Rute | CC BY-SA 3.0 |