Timeline for Dense subsets on set space
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2014 at 6:02 | comment | added | Michael Greinecker | @NateEldredge No disagreement there. | |
| Apr 30, 2014 at 13:22 | comment | added | Nate Eldredge | @Michael: Unless I am misreading, it gives a necessary and sufficient condition on a measure space for $L^p$ to be separable. I agree that one still has to investigate whether those conditions could fail for a metric space with a probability measure. It doesn't answer the question but it seemed relevant. | |
| Apr 30, 2014 at 7:36 | comment | added | Michael Greinecker | @NateEldredge At that link, there is no discussion as to whether probability measures on nonseparable metric spaces can be nonseparable. | |
| Apr 30, 2014 at 1:02 | comment | added | Nate Eldredge | See the link in my previous comment. | |
| Apr 30, 2014 at 0:55 | comment | added | Yee Neil | Would you like to give some more detail? I know that $(X,\mathscr{B})$ is separable if and only if the space $L^{1}(X,\mathscr{B},m)$ is separable ,but I donot know when $L^{1}(X,\mathscr{B},m)$ is separable. | |
| Apr 29, 2014 at 21:37 | comment | added | Nate Eldredge | Note that by identifying sets with their indicator functions, your metric space $(\mathcal{B},d)$ embeds as a total subset of $L^1(X,m)$. So I think your question is equivalent to asking about the separability of $L^1(X,m)$ where $X$ is metrizable. The answers to this question seem to apply here. | |
| Apr 29, 2014 at 16:47 | answer | added | Michael Greinecker | timeline score: 2 | |
| Apr 29, 2014 at 15:56 | comment | added | Ramiro de la Vega | @Michael: why don´t you turn your comment into an answer? | |
| Apr 29, 2014 at 13:22 | review | Suggested edits | |||
| Apr 29, 2014 at 13:33 | |||||
| Apr 29, 2014 at 8:10 | history | edited | András Bátkai | CC BY-SA 3.0 |
added arxiv tag
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| Apr 29, 2014 at 7:42 | comment | added | Michael Greinecker | Even after identifying equivalence classes, the answer might be no. If $\mathfrak{c}$ is a real-valued measurable cardinal, we can use the discrete metric and put an atomless probability measure on all subsets. The resulting measure will be as far as possible from separable by the Gitik-Shelah theorem. But the answer will be yes if the metric space is separable. | |
| Apr 29, 2014 at 7:28 | comment | added | Martin Sleziak | I don't think this is a metric: $d([0,1],(0,1))=0$, if we work with the usual Lebesgue measure on $\mathbb R$. Do you identify the sets such that $m(A\triangle B)=0$? (So the space would be the space of equivalence classes rather than the space of all Borel sets?) | |
| Apr 29, 2014 at 7:20 | history | asked | Yee Neil | CC BY-SA 3.0 |