On the complex plane $\mathbb C$ consider the half-open square $$\square=\{z\in\mathbb C:0\le\Re(z)<1,\;0\le\Im(z)<1\}.$$

Observe that for every $z\in \mathbb C$ and $p\in\{0,1,2,3\}$ the set $(z+i^p\cdot\square)$ is the shifted and rotated square $\square$ with a vertex at $z$.

Problem. Is it true that for any function $p:\mathbb C\to\{0,1,2,3\}$ there a subset $Z\subset\mathbb C$ such that the union of the squares $$\bigcup_{z\in Z}(z+i^{p(z)}\cdot\square)$$is not Borel in $\mathbb C$?

Added in Edit. As @YCor observed in his comment, the answer to this problem is affirmative under $\neg CH$.

An affirmative answer to Problem would follow from an affirmative answer to another intriguing

Problem'. Is it true that for any partition $\mathbb C=A\cup B$ either $A$ contains an uncountable strictly increasing function or $B$ contains an uncountable strictly decreasing function?

Here by a function I understand a subset $f\subset \mathbb C$ such that for any $x\in\mathbb R$ the set $f(x)=\{y\in\mathbb R:x+iy\in f\}$ contains at most one element.

Added in the Next Edit. In the discussion with @YCor we came to the conclusion that under CH the answer to both problems is negative. Therefore, both problems are independent of ZFC. Very strange.

  • $\begingroup$ It's even "a non-Borel union of unit squares". $\endgroup$ – YCor May 1 at 19:42
  • $\begingroup$ In view of my answer, I'm wondering if you wanted for $Z$ to be Borel. $\endgroup$ – Nate Eldredge May 1 at 19:43
  • $\begingroup$ @NateEldredge it should be very easy to adapt your construction if $Z$ is assumed to be Borel but not $p$. $\endgroup$ – YCor May 1 at 19:56
  • 1
    $\begingroup$ If for some line $L$ of negative slope we have $p^{-1}(\{0,2\})\cap L$ uncountable, then it contains a non-Borel subset and Nate's argument adapts. Idem if for some line $L$ of positive slope we have $p^{-1}(\{1,3\})\cap L$ uncountable. If none applies (for lines of slope $\pm 1$), CH holds, so already we're done in the negation of CH. $\endgroup$ – YCor May 1 at 20:08
  • 1
    $\begingroup$ But under CH I'd be surprised if there would exist a subset of the plane meeting every affine line of negative slope into a countable subset and with complement meeting every affine line of positive slope into a countable subset. $\endgroup$ – YCor May 1 at 20:10

(This addresses a misinterpretation of the question, where $p$ can be chosen. I'll try to fix it.)

This seems too easy, so maybe I've misunderstood the question, but: let $L$ be the diagonal line $\{z : \Re(z) = - \Im(z)\}$ and let $Z$ be a non-Borel subset of $L$. Take $p \equiv 0$. Then the set in question is $E = \bigcup_{z \in Z} (z + \Box)$, but we have $E \cap L = Z$ so that $E$ is not Borel.

  • $\begingroup$ It is not that easy: note that we have the function $p$, which rotates the squares. It may happen that on this diagonal line $p(z)=1$ then the union of such squares will be the open strip. $\endgroup$ – Taras Banakh May 1 at 19:51
  • $\begingroup$ Remark: assuming (as we can) that $Z$ is dense in $L$, then $E$ is equal to the open strip $\{x+iy:0<x+y<2\}$ union $Z$. $\endgroup$ – YCor May 1 at 19:53
  • $\begingroup$ @TarasBanakh I don't know if you're addressing Nate's answer or my comment, but in both we have $p=0$ as assumption. $\endgroup$ – YCor May 1 at 19:54
  • 1
    $\begingroup$ @YCor The problem was to find a non-Borel union for ANY function $p$, not for SOME $p$. $\endgroup$ – Taras Banakh May 1 at 19:55

Both problems have negative answer under CH and positive answer under $\neg$CH. The proofs can be found here.


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.