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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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(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.

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

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