5
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
3
+50
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.