Timeline for Existence of measurable "inclusion" into Euclidean space
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2020 at 0:03 | history | became hot network question | |||
| May 13, 2020 at 19:53 | vote | accept | ABIM | ||
| May 13, 2020 at 19:23 | answer | added | Nate Eldredge | timeline score: 4 | |
| May 13, 2020 at 19:13 | answer | added | Michael Greinecker | timeline score: 4 | |
| May 13, 2020 at 19:12 | comment | added | Nate Eldredge | @AIM_BLB: Yes. A submetrizable space, by definition, has a continuous injection into a metric space, which in turn has a continuous injection into its completion, which as abx noted has a bimeasurable bijection to $\mathbb{R}$ assuming separability in the right places. | |
| May 13, 2020 at 18:32 | comment | added | ABIM | @NateEldredge I never heard of submetrizable, but I assume it can be worked out directly from the theorem abx provided? | |
| May 13, 2020 at 18:32 | comment | added | ABIM | @abx I did not know this result, thanks. | |
| May 13, 2020 at 18:20 | comment | added | abx | @Nate Eldredge: Yes, right. | |
| May 13, 2020 at 17:40 | comment | added | Nate Eldredge | @abx: Since only injectivity is requested here, any separable metrizable space will do. In particular, a non-measurable subset of $\mathbb{R}^n$ with its Borel $\sigma$-algebra will do. Actually, submetrizable is enough. | |
| May 13, 2020 at 17:03 | comment | added | YCor | In general, we can stick to $n=1$. | |
| May 13, 2020 at 16:59 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title; edited tags
|
| May 13, 2020 at 16:33 | comment | added | abx | Such a function (in fact, bijective) exists as soon as $\Omega $ is an uncountable polish space — this is the Borel isomorphism theorem. | |
| May 13, 2020 at 16:03 | history | asked | ABIM | CC BY-SA 4.0 |