Partitioning $\mathbb{R}^3$ into two convex sets.

Question

I know that there are $4$ partitions of $\mathbb{R}^3$ into two convex sets, up to affine transformation. One of the partitions is the trivial $\{\emptyset,\mathbb{R}^3\}$ (note that by partition, I am requiring an unordered pair of disjoint sets whose union is $\mathbb{R}^3$). Unfortunately I'm having a hard time figuring out the other partitions. Geometrically, I'm quite lost on how this is possible. If anyone could help give some intuition on this idea, that would be greatly appreciated. I am not looking for any kind of rigorous proof.

Answer 1

For the others, we cut at a hyperplane, then cut that hyperplane up into two convex parts and assign one part to each half.

Now, cutting the plane. Our first option is to split it as all versus none. The alternative is to cut it with a line, then split that line into two convex parts and assign one part to each half.

We can split the line in two ways: all versus none, or split at a point. That point is then assigned to one of the parts.

That's four options. Examples of one of the convex parts for each option:
$$\mathbb{R}^3$$ $$\{(x,y,z): x\ge 0\}$$ $$\{(x,y,z): (x > 0)\text{ or }(x = 0\text{ and }y\ge 0)\}$$ $$\{(x,y,z): (x > 0)\text{ or }(x = 0\text{ and }y > 0)\text{ or }(x=0\text{ and }y=0\text{ and }z\ge 0)\}$$ In $d$ dimensions, this naturally extends to $d+1$ ways to split the space.