Timeline for Is definability of a basis for $\mathbb{R^N}$ independent of ZFC?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2013 at 23:28 | vote | accept | Colin McLarty | ||
| Nov 19, 2013 at 3:00 | comment | added | Sam Hopkins | @Colin: perhaps you could edit either the title or content of your post to make the way you ask your question consistent, because it makes it hard to parse the answers you have been getting otherwise. | |
| Nov 18, 2013 at 21:13 | answer | added | Asaf Karagila♦ | timeline score: 4 | |
| Nov 18, 2013 at 20:41 | comment | added | Joel David Hamkins | Yes, Monroe, you are right; I had also thought of adding $\omega_1$ many Cohen reals, but couldn't prove it in that case either. | |
| Nov 18, 2013 at 20:40 | comment | added | Asaf Karagila♦ | @Joel: First of all, one can use Shelah's model of $\sf ZFC+BP^{HOD(\Bbb R)}$ and eliminate the inaccessible. Alas, my argument is that a definable function is immediately Baire measurable, where I can only conclude that it is Baire measurable with respect to definable subsets, not in general. So I'm still not sure, but there is probably some fine point I'm missing right now which can complete the argument. | |
| Nov 18, 2013 at 20:25 | comment | added | Monroe Eskew | @Joel: If you start with $L$ and add one generic real $r$, then $HOD(\mathbb{R})^{V[r]} = L[r]$, so there is a basis in this model. Maybe $\omega_1$ Cohen reals is enough. | |
| Nov 18, 2013 at 20:03 | comment | added | Asaf Karagila♦ | @Joel: I don't know, I didn't really think it through when I wrote the comment. But you're right. I'll post an answer with some extended references, and we can continue the discussion from there. | |
| Nov 18, 2013 at 19:49 | comment | added | Joel David Hamkins | But why do you require it to be projective, rather than just full ordinary definability? After all, every definable subset of $\mathbb{R}^{\mathbb{N}}$ is in $\text{HOD}(\mathbb{R})$. I don't see the argument you have in mind, but if what you say is right, then you've got an answer to Colin's question (modulo an inaccessible cardinal). Namely, a model of ZFC with no definable basis for $\mathbb{R}^{\mathbb{N}}$. | |
| Nov 18, 2013 at 19:43 | comment | added | Asaf Karagila♦ | @Joel: The argument which eludes you is an interesting problem. It seems to be closely related to a long standing problem of whether or not there is a Hamel basis for $\Bbb R$ over $\Bbb Q$ in Cohen's first model (there are Vitali sets there, by the way). It is easy to show, though, that collapsing an inaccessible to be $\aleph_1$ then there is no basis for $\Bbb{R^N}$ in $L(\Bbb R)$, or $\sf HOD(\Bbb R)$ (by standard Solovay model arguments), and therefore there is no definable basis if we require definable to be projective, or so, in the full universe. | |
| Nov 18, 2013 at 19:37 | comment | added | Joel David Hamkins | Meanwhile, I believe that if one adds a Cohen real, then there will be no definable basis for $\mathbb{R}^{\mathbb{N}}$ in $V[c]$, because I think even that there is no basis at all in $\text{HOD}(\mathbb{R})^{V[c]}$. But verifying the details of this argument elude me... | |
| Nov 18, 2013 at 19:36 | comment | added | Joel David Hamkins | Peter, the OP already has your situation (b), since under V=L, there is the formula defining membership in the L-least basis, which consistently is a basis. More generally, this works whenever there is a definable well-ordering of the reals. Indeed, any model of ZFC can be extended to a forcing extension, not adding reals, in which there is a definable basis. | |
| Nov 18, 2013 at 18:56 | comment | added | Peter LeFanu Lumsdaine | Just to ward off ambiguity: by a “definable basis in ZFC”, do you mean (a) “a formula which (provably in ZFC) defines a basis for $\mathbb{R}^\mathbb{N}$”, or (b) “a formula which (provably in ZFC) defines a unique set, which (consistently with ZFC) is a basis for $\mathbb{R}^\mathbb{N}$”? (If I remember right, I’ve seen people mean both of these things when asking “can we find a definable $X$ with property $P$ in $T$?” here.) | |
| Nov 18, 2013 at 18:28 | comment | added | Asaf Karagila♦ | Define "definable"? It is consistent with $\sf ZF$ that there is no basis to $\Bbb{R^N}$, which may be an indication - in this context - that the answer is negative. | |
| Nov 18, 2013 at 17:48 | history | asked | Colin McLarty | CC BY-SA 3.0 |