A non-Borel union of unit half-open squares

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.

• It's even "a non-Borel union of unit squares". – YCor May 1 at 19:42
• In view of my answer, I'm wondering if you wanted for $Z$ to be Borel. – Nate Eldredge May 1 at 19:43
• @NateEldredge it should be very easy to adapt your construction if $Z$ is assumed to be Borel but not $p$. – YCor May 1 at 19:56
• 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. – YCor May 1 at 20:08
• 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. – 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.
• 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. – Taras Banakh May 1 at 19:51
• 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 $Z$. – YCor May 1 at 19:53
• @TarasBanakh I don't know if you're addressing Nate's answer or my comment, but in both we have $p=0$ as assumption. – YCor May 1 at 19:54
• @YCor The problem was to find a non-Borel union for ANY function $p$, not for SOME $p$. – 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.