Write in a concise form a set in 4-dimensional space.

Question

I have the following subset of the 4-dimensional space $(x_1,x_2,x_3,y)$: \begin{align} \mathcal I=&\left\{x_1=0\, , \, y\in[0, 1]\, , \, 0 \leq x_2\leq y\, , \, 0 \leq x_3\leq y\right\}\cup\\ &\left\{x_2=0\, , \, y\in[0, 1]\, , \, 0 \leq x_1\leq y\, , \, 0 \leq x_3\leq y\right\}\cup\\ &\left\{x_3=0\, , \, y\in[0, 1]\, , \, 0 \leq x_1\leq y\, , \, 0 \leq x_2\leq y\right\}\cup\\ &\left\{y=x_1\, , \, x_1\in[0,1] \, , \, 0 \leq x_2\leq x_1\, , \, 0 \leq x_3\leq x_1\right\} \cup\\ &\left\{y=x_2\, , \, x_2\in[0,1] \, , \, 0\leq x_1\leq x_2\, , \, 0 \leq x_3\leq x_2\right\} \cup\\ &\left\{y=x_3\, , \, x_3\in[0,1] \, , \, 0 \leq x_1\leq x_3\, , \, 0 \leq x_2\leq x_3\right\} \end{align} I would like to write $\mathcal I$ in a more concise form. Do you think it's possible? For example using a Cartesian product of sets.

Answer 1

How about $$ {\cal{I}}=\{(x_1,x_2,x_3,y)\in[0,1]^4 : \max_i x_i \le y \;\wedge\; x_1x_2x_3(y-x_1)(y-x_2)(y-x_3)=0 \} $$ ?

Answer 2

Let's reduce the problem by one dimension, fixing $x_3 = 0$. And to make it a little easier to wrap my head around things, I'm going to rename your variables $x_0, \ldots, x_3$ to get $$\mathcal I=\\\left\{x_0=0\, , \, x_1 \in[0, 1]\, , \, 0 \leq x_2\leq x_1\, , \, 0 \leq x_3\leq x_1\right\}\cup\\ \left\{x_2=0\, , \, x_1\in[0, 1]\, , \, 0 \leq x_0\leq x_1\, , \, 0 \leq x_3\leq x_1\right\}\cup\\ \left\{x_3=0\, , \, x_1\in[0, 1]\, , \, 0 \leq x_0\leq x_1\, , \, 0 \leq x_2\leq x_1\right\}\cup\\ \left\{x_1=x_0\, , \, x_0\in[0,1] \, , \, 0 \leq x_2\leq x_0\, , \, 0 \leq x_3\leq x_0\right\} \cup\\ \left\{x_1=x_2\, , \, x_2\in[0,1] \, , \, 0\leq x_0\leq x_2\, , \, 0 \leq x_3\leq x_2\right\}\cup\\ \left\{x_1=x_3\, , \, x_3\in[0,1] \, , \, 0 \leq x_0\leq x_3\, , \, 0 \leq x_2\leq x_3\right\} $$

Now restricting to one lower dimension simplifies this to the analogous problem $$ \mathcal I=\\ \left\{x_0=0, x_1 \in[0, 1], 0 \leq x_2\leq x_1 \right\}\cup\\ \left\{x_2=0, x_1\in[0, 1], 0 \leq x_0\leq x_1\right\}\cup\\ \left\{x_1=x_0, x_0\in[0,1], 0 \leq x_2\leq x_0 \right\}\cup\\ \left\{x_1=x_2, x_2\in[0,1], 0\leq x_0\leq x_2\right\} $$ And substituting the more usual variable names gives $$ \mathcal I=\\ \left\{x=0, y \in[0, 1], 0 \leq z\leq y \right\}\cup\\ \left\{z=0, y\in[0, 1], 0 \leq x\leq y\right\}\cup\\ \left\{y=x, x\in[0,1], 0 \leq z\leq x \right\}\cup\\ \left\{y=z, z\in[0,1], 0\leq x\leq z\right\} $$

When you look at that, each of the four sets is a triangle, the first two lying in the $yz$ and $xz$ coordinate planes, the last two lying in the planes defined by $y = x$ and $z = x$. Calling the four triangles $A, B, C, D$ In order, we see that $A$ and $C$ share only the origin; the same goes for $B$ and $D$. $A$ and $B$ meet along a line segment consisting of points $(0,t,0)$ where $0 \le t \le 1$.

Continuing in this way and then doing a little plotting, you find that $B$ and $C$ share an edge, $C$ and $D$ share a different edge, and $D$ and $A$ share an edge. You end up with something that's topologically just a fan made of four triangles with a shared vertex. But there's no hope of describing it as a union of fewer than four planar pieces; for instance, any cartesian product of intervals must be planar, and the four planar pieces in our set all lie in different planes. In short, there's no simpler way I can think of writing this as you've described.

There is a slight simplification that's possible. Consider the four-segment path, with vertices $$ (1,0,0)\\ (1,1,0)\\ (1,1,1)\\ (0,1,1) $$

The "cone" on this path (with the origin as vertex) ends up being the union of the four triangles. (The "cone" is the union of the segments $OP$ over all points $P$ in the path.)

In your 4D example, something similar is true: a collection of six triangles forms a "surface" in 4D (I think it's a "strip", i.e., a sequence of triangles where only adjacent triangles in the sequence meet, and they meet along a common edge), and the cone on this strip is a solid that you've called $I$.