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The problem is to prove the following equivalence:

$$2^{\aleph_0}=\aleph_1 \iff \exists\ A,B\in\mathbb{R}^2:\\ \textrm{a)}A\cup B=\mathbb{R}^2 \\\textrm{b)}\ \forall\ \textrm{lines } l\ \textrm{parallel to x-axis, } |A\cap l|\leq\aleph_0 \\\textrm{c)}\ \forall\ \textrm{lines } l\ \textrm{parallel to y-axis, } |B\cap l|\leq\aleph_0$$

I think I know how to prove "$\impliedby$":

Let's assume the RHS and that $2^{\aleph_0}\geq\aleph_2$. Now let's take a sequence $\langle x_{\alpha}:\ \alpha<\omega_1\rangle$. Since $\bigcup_\limits{\alpha<\omega_1}|B\cap x_{\alpha}|\leq \aleph_1\cdot\aleph_0=\aleph_1$ there is such a $y$-coordinate that neither of $(x_{\alpha},y)$ is in $B$. But hence we obtain $\aleph_1$ point which all lie on the same line parallel to $x$-axis. Since $A\cup B=\mathbb{R}^2$ they all must be contained in $A$ $-$ a contradiction as $A$ has at most countably many of those points.

But what about the other implication? Do I contruct these sets by induction? Over all lines or something else? I haven't come up with anything yet, perhaps you could help me.

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Hint. Let $\sqsubseteq$ well-order $\mathbb R$ of order type $\omega_1$. Being a relation, $\sqsubseteq$ is a subset of $\mathbb R\times\mathbb R$. Perhaps that will work as either $A$ or $B$?

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  • $\begingroup$ Thanks for the hint! I need to think about it a moment :D $\endgroup$
    – Jules
    Jul 27, 2016 at 8:40
  • $\begingroup$ I still don't get it :( Maybe a hint to a hint? I mean: Every linear order contains a cofinal well-order. So what if $\sqsubseteq$ is a cofinal well order on some line $l$ with natural order? $\endgroup$
    – Jules
    Jul 27, 2016 at 9:58
  • $\begingroup$ @Jules: You're overcomplicating it :). Assuming CH, there is a bijection $\varphi:\mathbb R\to\omega_1$. Then define $x\sqsubseteq y$ iff $\varphi(x)\le\varphi(y)$. This gives you a well-ordering of all of $\mathbb R$ of order type $\omega_1$. In particular $\{x\in\mathbb R\mid x\sqsubseteq a\}$ is at most countable for every $a$. $\endgroup$
    – hmakholm left over Monica
    Jul 27, 2016 at 10:09
  • $\begingroup$ @Jules: Remember what an ordering is. An ordering such as $\sqsubseteq$ is a particular case of a relation, which is, in and of itself, a set of pairs, that is, a subset of $\mathbb R\times\mathbb R$. The set $\sqsubseteq$ itself can be your $A$. And then $B$ becomes $\mathbb R^2 \setminus {\sqsubseteq}$. $\endgroup$
    – hmakholm left over Monica
    Jul 27, 2016 at 10:33
  • $\begingroup$ But don't I need a set on which I impose this order? A set of elements that this relation pertains to? I guess it should be $\mathbb{R}^2$ here. $\endgroup$
    – Jules
    Jul 27, 2016 at 10:37

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