Restrictions of null/meager ideal

Question

Let I denote the null (resp. meager) ideal on reals. Is it consistent that for any pair of non null (resp. meager) sets A and B, there is a null (resp. meager) preserving bijection between A and B? In particular, is this true in the model obtained by adding $\omega_2$ Cohen (resp. random reals) over a model of CH?

Answer 1

$\newcommand\continuum{\mathfrak{c}}$

Update. This proof strategy is hopeless, for the reasons explained in Miha's answer at mathoverflow.net/a/144538/1946, and the comments on it. But I'll leave it here for the record of this false attempt.

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Original answer:

Your hypothesis follows from the continuum hypothesis, and more generally, from the assertion that the additivity of the null ideal is the continuum, which is consistent with the failure of CH.

To see this, suppose that the additivity of the null ideal is the continuum, which means that the union of fewer than continuum many sets of measure zero still always has measure zero, and in particular, every set of size less than the continuum has measure zero. Fix any two sets $A$ and $B$ that are not of measure zero. So they have size continuum. Let us now build a bijection between them $\pi:A\to B$, in such a way that $\pi$ takes every measure zero subset of $A$ to a measure-zero subset of $B$ and conversely. We will construct $\pi$ as the union of a increasing chain of partial functions $\pi=\bigcup_{\alpha\lt\continuum}\pi_\alpha$, in a construction of length continuum, where $\pi_\alpha:A_\alpha\to B_\alpha$ is a bijection of measure-zero subsets of $A$ and $B$, respectively. Enumerate the Borel measure-zero sets as $D_\alpha$ for $\alpha\lt\continuum$. Suppose that $\pi_\alpha:A_\alpha\to B_\alpha$ is defined, and consider the set $D_\alpha$. First, we may extend $A_\alpha$ to $A_{\alpha+1}$ in such a way that $D_\alpha\cap A\subset A_{\alpha+1}$, by also adding a measure zero part of $B$ to $B_\alpha$, forming $B_{\alpha+1}$ and extending the bijection to $\pi_{\alpha+1}:A_{\alpha+1}\to B_{\alpha+1}$. (I am using that every non-measure-zero set contains a measure zero set of size continuum, which I believe follows from our assumptions; please correct me if this is wrong. Edit: this is not correct, because of the possibility of Sierpinski and Luzin sets, which exist, as Miha pointed out, when the additivity numbers are large.) Similarly, for the target side, we may extend in such a way also that $D_\alpha\cap B\subset B_{\alpha+1}$. At limit stages of our construction, we take the union of the bijections constructed so far, and until the end of the constructin, this union domain and target will still have measure zero by the assumption that the additivity of the null ideal is the continuum.

The resulting map $\pi:A\to B$ is a bijection, since every point will arise as a singleton in some $D_\alpha$, thereby getting added to the domain and range at that stage. And every measure zero subset of $A$ will be covered by some $A\cap D_\alpha$ for some $\alpha$, which gets mapped to $B_{\alpha+1}$, which has measure zero on the $B$ side. And similarly for the measure zero subsets of $B$.

Thus, the construction resembles the familiar back-and-forth constructions, but at each stage, we have defined the bijection on only measure zero sets, which exhaust neither $A$ nor $B$. This gives time to diagonalize against the possible measure zero sets, since there are only continuum many Borel measure zero sets to consider.

So the answer is yes, this situation is consistent.

Meanwhile, if the uniformity number (the smallest size of the non-measure zero set) is less than the continuum, then your hypothesis clearly fails, since there will be non-measure-zero sets of different cardinality, which can therefore not be bijective. So we seem to have trapped your statement between the assertions that the additivity number is big and the uniformity number is not small.

A very similar argument seems to work also in the case of category rather than measure, under the assumption that the additivity of the meager ideal is the continuum. (And one similarly needs in this case that every non-meager set contains a meager set of size continuum.)