By $\mathbb{R^N}$ I mean the real vector space with the natural componentwise addition and scalar multiplication. Certainly ZFC+(V=L) gives definable bases, but does ZFC?
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1$\begingroup$ 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. $\endgroup$– Joel David HamkinsCommented Nov 18, 2013 at 19:36
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1$\begingroup$ 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... $\endgroup$– Joel David HamkinsCommented Nov 18, 2013 at 19:37
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1$\begingroup$ @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. $\endgroup$– Asaf Karagila ♦Commented Nov 18, 2013 at 19:43
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1$\begingroup$ 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 $\mathbb{R}^{\mathbb{N}}$. $\endgroup$– Joel David HamkinsCommented Nov 18, 2013 at 19:49
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1$\begingroup$ @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. $\endgroup$– Monroe EskewCommented Nov 18, 2013 at 20:25
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The answer, if I understand the question correctly, is negative. That is, if we understand "definable" as "can be defined from ordinals [and a real number]", or in simpler words, sets which are in $\sf HOD(\Bbb R)$. Here is a sketch of an argument I suspect is correct.
Consider Solovay's model which we get from collapsing an inaccessible cardinal to $\aleph_1$. In that model every set of real numbers definable from an ordinal and a real number has the Baire property. Since I am not going to care about the forcing, $V$ is going to be the model after the collapse.
Now suppose that $B$ was a basis and was definable, since $\sf HOD(\Bbb R)$ and $V$ both agree that $B$ must have the cardinality of the real numbers, they also agree that it has more than $2^{\aleph_0}$ permutations. Each of those permutations extends uniquely to an automorphism of $\Bbb{R^N}$. Therefore in $\sf HOD(\Bbb R)$ there are more than $2^{\aleph_0}$ automorphisms of the space.
However in $\sf HOD(\Bbb R)$ we have automatic continuity for Polish groups, and therefore every automorphism is continuous. This is a contradiction since there can only be $2^{\aleph_0}$ continuous automorphisms of a Polish space.
Therefore $B$ cannot be definable to begin with. One can also get away from the inaccessible cardinal by considering Shelah's model in which not all sets are Lebesgue measurable, but all sets have the Baire property.
Solovay, Robert M., "A model of set-theory in which every set of reals is Lebesgue measurable." Ann. of Math. (2) 92 1970 1–56.
Shelah, Saharon, "Can you take Solovay's inaccessible away?" Israel J. Math. 48 (1984), no. 1, 1–47.
Raisonnier, Jean, "A mathematical proof of S. Shelah's theorem on the measure problem and related results." Israel J. Math. 48 (1984), no. 1, 48–56.
answered Nov 18, 2013 at 21:13
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$\begingroup$ I am learning about this from reading the comments. But it seems to me that being in $\sf HOD(\Bbb R)$ is at least as broad a sense of definable as I would want to know about. $\endgroup$ Commented Nov 19, 2013 at 16:51
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$\begingroup$ Colin, do note that under "reasonable" assumptions (e.g. that $V$ is the collapse of an inaccessible cardinal) this model does not satisfy the axiom of choice. However my argument is that essentially being a basis of a definable vector space is to some extent absolute. So $B$ is a basis in $\sf HOD(\Bbb R)$ if and only if it is a basis in $V$. $\endgroup$– Asaf Karagila ♦Commented Nov 19, 2013 at 16:54
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$\begingroup$ You say "to some extent absolute," but I believe you mean it is rigorously true that $B$ is a basis in $V$ iff it is in $V$. Right? I think the point is that $B$ being a basis is a matter relating individual elements of $V$ to finite subsets of $B$ and the field. Is that right? Or at least reasonably well stated? $\endgroup$ Commented Nov 20, 2013 at 13:23
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$\begingroup$ Colin, yes. Since the structure of the vector space is certainly in $\sf HOD(\Bbb R)$, we have that being a basis is something which depends on the finite subsets which are certainly absolute to that inner model. $\endgroup$– Asaf Karagila ♦Commented Nov 20, 2013 at 13:54