Timeline for Borel set plus a closed set = Borel
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2010 at 18:53 | comment | added | Andrés E. Caicedo | It is a source of confusion, since in descriptive set theory most results are independent of the specific Polish space we use. I guess I could add it to the list of common mistakes... | |
| Dec 7, 2010 at 18:16 | comment | added | gowers | Ah yes, a closed subset of $\mathbb{R}^2$ can be written as a union of countably many compact sets, so it obviously can't work. | |
| Dec 7, 2010 at 17:53 | comment | added | Andrés E. Caicedo | @gowers : Yes. The general result in ${\mathbb N}^{\mathbb N}$ that projections of closed sets are not Borel does not transfer intact to ${\mathbb R}$. | |
| Dec 7, 2010 at 17:43 | comment | added | gowers | Are you saying that a projection of a closed set in $\mathbb{R}^2$ has to be Borel? | |
| Dec 7, 2010 at 16:56 | comment | added | Andrés E. Caicedo | @gowers : Actually, I do not think this is correct. (It is if instead of the reals we work with the irrationals.) To get non-Borel sets it suffices to project the complement (in ${\mathbb R}^2$) of the projection of a closed subset of ${\mathbb R}^3$. | |
| Dec 7, 2010 at 16:55 | history | edited | gowers | CC BY-SA 2.5 |
added 243 characters in body
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| Dec 7, 2010 at 16:25 | history | answered | gowers | CC BY-SA 2.5 |