Building $\mathbb R^2$ as a countable union of bi-partitions products.

Question

Let the set $S$ be initially empty.

In one step you can partition $\mathbb R$ into two sets $A, B$ and add $A\times B$ to $S$.

Can you make so that $S = \mathbb R^2\setminus\{(x,x):x\in\mathbb R\}$ in countably many steps?

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Of course, there is nothing special about $\mathbb R$ but it's cardinality in this problem, so feel free to change it to $\mathcal P(\mathbb N)$ or $\{0, 1\}^\mathbb N$.

My guess is that it is possible. No clue on how to approach it, though.

I don't really have a motivation for this one. I just have been questioning a lot of things about $\mathbb R$ lately.

Answer 1

For every rational number $q$, add $(-\infty,q] \times (q,\infty)$ and its mirror image $(q,\infty) \times (-\infty,q]$ to $S$.

This does the job:

• For every point $(x,y)$ with $x<y$, there is a rational number $q$ between $x$ and $y$; then $(-\infty,q] \times (q,\infty)$ contains $(x,y)$.
• Similarly, for every point $(x,y)$ with $x>y$, we can take a rational number $q$ between $y$ and $x$, and see that $(q,\infty) \times (-\infty,q]$ contains $(x,y)$.
• But of course none of the sets we add to $S$ contain any point of the form $(x,x)$, so their union is precisely the desired set $\{(x,y) \in \mathbb R^2 : x \ne y\}$.