Is there a crossing-free planar embedding of the 2-skeleton of the 6-simplex?

Question

A pseudo-disk arrangement is a collection of planar bodies whose boundaries are Jordan curves that pairwise intersect at most twice. I would like to know if given seven points in the plane whether it is possible to find a pseudo-disk arrangement with $\binom 73$ disks, such that for each triple of the seven points there is one disk containing exactly them from the set of seven points. I'm also interested in what kind of theorems there are to conclude the non-existence of such embeddings.

I've just realized that I don't even know the answer if we replace seven with five. For five points we can find a pseudo-ball arrangement is $3d$; take a tetrahedron with a point inside, and slightly perturb the ten triangles they span. Can we do more points in $3d$? In $4d$ any number of points are OK if we aim for triples (just like pairs would be OK in $3d$).

Please note that I've updated the question a couple of times to make the problem more clear, some of the comments and Jeff's answer are for older versions.

Answer 1

I think that there is no pseudo-disk arrangement even with $\binom{5}{3}$ pseudo-disks for the 5 point case; I sketch a proof below.

Suppose for contradiction that there is such an arrangement. Let $a_1,\dots, a_5$ be the points. For each closed Jordan curve $c_{ijk}$ that surrounds $a_i,a_j$ and $a_k$ we define a Jordan arc $d_{lm}$ that connects the remaining two points $a_l$ and $a_m$ in the unbounded region of $c_{ijk}$. We should define the $d$-curves in a way that there are no touchings between them or with the $c$-curves, and also we should make sure that the number of curve incidences in the whole picture is finite.

I claim that a pair of $d$-curves that don't share an endpoint must intersect an even number of times Wlog. consider the curves $d_{12}$ and $d_{45}$. The closed Jordan curves $c_{123}$ and $c_{345}$ together define four regions in the plane, one unbounded, one containing $a_1$ and $a_2$, one containing $a_3$, and one containing $a_4$ and $a_5$. Let $I$ be the union of the three bounded regions. Notice that the boundary of $I$ is made up of two curves, one is an arc of $c_{123}$, while the other is an arc of $c_{345}$.

Since any intersection point $q \in d_{12} \cap d_{45}$ must lie outside $I$, both $d_{12}$ and $d_{45}$ has at least one arc outside $I$. Consider the space $\mathbb{R}^2/I$. In this space, the curves $d_{12}$ and $d_{45}$ become collections of closed Jordan curves that contain the point $p$ that is the picture of $I$ in the contraction. Since one boundary arc of $I$ is only intersected by $d_{12}$ and the other only by $d_{45}$, the curve endigns around $p$ are `separable', i.e., if we label each curve ending by $a$ for the curve $d_{12}$ and by $b$ for the curve $d_{45}$, then the cyclic order of the labels around $p$ is $a^{2k}b^{2l}$ for some positive integers $k,l$.

Let $g_1,\dots,g_k$ be the closed Jordan curves of $d_{12}/I$ and let $g'_1,\dots,g'_l$ be the closed Jordan curves of $d_{45}/I$. In order to show that $|d_{12} \cap d_{45}|$ is even, it is sufficient to show that $|g_i \cap g'_j|$ is even for any pair $i,j$. We have seen that $g_i$ and $g'_j$ are two closed Jordan curves that touch at the singular point $p$. It follows that they must have an even number of crossings.

The $d$-curves give a planar drawing of the 5-clique, where the curves corresponding to any pair of non-incident edges intersect an even number of times. This contradicts the strong Hanani-Tutte theorem.

Answer 2

Here is a positive answer using connected sets whose boundary is not a Jordan curve.

We want $m={n \choose k}$ connected sets, $S_1, S_2, \ldots , S_m\subseteq \mathbb{R}^2$. I'll want to associate each of these sets with a $k$-tuple, so I'll just refer to the $i^\mathrm{th}$ $k$-tuple.

Start with the graphs $G_i$ of $f_i (x) = \sin({1\over x})+ {i\over 2m}$ for $x > 0$ and $i = 1, 2, \ldots, m$. Each of these graphs has the entire interval $I = \{ 0\} \times [{1\over 2}, 1]$ in its set of limit points. Break $I$ into $n$ subintervals $I_1, I_2, \ldots, I_n$ (ordered with increasing $y$-values.

Next, line up your $n$ points $a_1, a_2, \ldots, a_n$ along the line $x=-1$ (ordered with increasing $y$-values) and from $a_j$ draw ${n-1\choose 2}$ distinct line segments to the interval $I_j$.

Now for a the $i^\mathrm{th}$ $k$-tuple $\{ a_1, a_2, ..., a_k\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the $k$ segments

• out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3, \ldots, a_k\}$
• $\vdots$
• out of $a_k$ corresponding to $2$-tuple $\{a_1, a_2, \ldots, a_{k-1}\}$

The fact that these sets are connected is the standard use of the topologist's sine curve. They plainly do not intersect.

The same method can be used to embed any countable hypergraph in $\mathbb{R}^2$.