Suppose that for each $\alpha < 2^\omega$, $f_\alpha:2^\omega \rightarrow 2^\omega$ is a bijection.

I want to know whether it's always possible to construct an $X\subseteq Y\subseteq 2^\omega$ such that:

  1. $X$ has measure 0.
  2. $Y$ has positive measure.
  3. $\{\alpha \mid \min f_\alpha(Y) = \min f_\alpha(X)\}$ has positive measure.

(For $Z\subseteq 2^\omega$, $\min Z$ denotes the smallest element under the ordinal ordering.)

  • $\begingroup$ Why use $\inf$ when well-ordering have $\min$? $\endgroup$ – Asaf Karagila Mar 15 '16 at 5:57
  • $\begingroup$ Yes, I meant to use $\min$. (I have edited the question, and also corrected an omission). $\endgroup$ – Andrew Bacon Mar 15 '16 at 6:05
  • $\begingroup$ In 2 and 3, do you also want these sets to be Lebesgue measurable, or do you just want them to have positive outer measure? (Measurability seems like it might be a pretty strict requirement in the case of 3.) $\endgroup$ – Paul McKenney Mar 15 '16 at 13:51
  • $\begingroup$ I did want the sets all to be measurable. (I would still be interested to know if the question holds if we relax that requirement to having positive outer measure though.) $\endgroup$ – Andrew Bacon Mar 15 '16 at 15:04

Here is a counterexample under CH. Let $\{x_i : i < \omega_1\}$ enumerate all reals. Let $\{B_i : i < \omega_1\}$ list all Borel null sets and put $N_i = \bigcup \{B_j : j < i\}$. Let $\langle K_i : i < \omega_1\rangle$ list all compact non null sets. For each $i < \omega_1$ choose a bijection $f_i$ satisfying:

(a) $f_i[N_i \setminus \{x_i\}] \subseteq \{x_j : j > i\}$

(b) $(\forall \omega \leq j < i)(f_i[K_j] \cap \{x_k : k < i\} \neq \phi)$

(c) $f_i(x_i) = x_i$

Now suppose $X$ is null and $Y$ is measurable non null. Choose $i_0$ such that $K_{i_0} \subseteq Y$ and $X$ is contained in $B_{i_0}$. Suppose $i > i_0$ and $x_i \notin X$. Then $\textsf{min}(f_i[Y]) \leq \textsf{min}(f_i[K_{i_0}]) < i$ by clause (b). Also $\textsf{min}(f_i[X]) \geq \textsf{min}(f_i[N_i] \setminus \{x_i\}) > i$. So $\{x_i : \textsf{min}(f_i[X]) = \textsf{min}(f_i[Y])\}$ is null.

I will try to say more on other variations soon.

  • $\begingroup$ Thanks! I'd be interested to hear more about the variations you mentioned if you get a chance to write those up at some point. Do you think it's likely that there are any axioms beyond ZFC that entail the opposite answer? $\endgroup$ – Andrew Bacon Mar 21 '16 at 16:35

Let us show that CH can be dropped from the previous argument (Check!). Let $\langle (N_i, K_i, x_i) : i < \mathfrak{c} \rangle$ satisfy the following: $\langle x_i : i < \mathfrak{c} \rangle$ is a one-one enumeration of all reals, for each Borel null set $N$ and Borel non null set $K$, the set $\{x_i : N_i = N, K_i = K\}$ is a Bernstein subset of reals (so it meets every perfect set of reals). Construct $\langle f_i : i < \mathfrak{c} \rangle$ such that the following hold

(a) $f_i[N_i \setminus \{x_i\}] \subseteq \{x_j : j > i\}$

(b) $f_i[K_i] \cap \{x_j : j < i\} \neq \phi$, for $i \geq \omega$

(c) $f_i(x_i) = x_i$

Suppose $X$ is null and $Y$ is measurable non null. Let $N$ be a null Borel set containing $X$. Let $K$ be a compact non null set contained in $Y$.

We claim that the set $W = \{x_i : \textsf{min}(f_i[X]) = \textsf{min}(f_i[Y])\}$ has zero inner measure. Suppose not. Then $W_1 = W \cap \{x_i : x_i \notin N, N = N_i, K = K_i\}$ is non null - Contradiction as before.

I don't have anything interesting to say on the "non null" case yet. I will post again when I have something.


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