Timeline for Non-Borel sets without axiom of choice
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2010 at 11:59 | comment | added | Joel David Hamkins | Gerald, yes, by Andres' answer, you need to use at least some choice to prove that there is any non-Borel set at all. In particular, in the model of ZF he mentions, Gowers' examples here are Borel, like the other examples we've had, because every set in that model is Borel. It appears that countable AC is sufficient to get the basic theory moving, and I expect that would be sufficient to prove that these examples are not Borel (but I haven't checked). | |
| Jul 22, 2010 at 11:53 | comment | added | Gerald Edgar | So are we to conclude from Andres Caicedo's answer that any PROOF that [the differentiable functions] is not Borel must use the Axiom of Choice? Probably, as remarked by Joel, we probably merely show the set is not $\Delta^1_1$ and then rely on AC to conclude it is not Borel. | |
| Jul 22, 2010 at 6:51 | history | answered | gowers | CC BY-SA 2.5 |