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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, , , yin[0, 1], , , 0 leq x_2leq y, , , 0 leq x_3leq yright}cup\ &left{x_2=0, , , yin[0, 1], , , 0 leq x_1leq y, , , 0 leq x_3leq yright}cup\ &left{x_3=0, , , yin[0, 1], , , 0 leq x_1leq y, , , 0 leq x_2leq yright}cup\ &left{y=x_1, , , x_1in[0,1] , , , 0 leq x_2leq x_1, , , 0 leq x_3leq x_1right}
cup\ &left{y=x_2, , , x_2in[0,1] , , , 0leq x_1leq x_2, , , 0 leq x_3leq x_2right}
cup\ &left{y=x_3, , , x_3in[0,1] , , , 0 leq x_1leq x_3, , , 0 leq x_2leq x_3right}
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.

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

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
$$mathcal I=\left{x_0=0, , , x_1 in[0, 1], , , 0 leq x_2leq x_1, , , 0 leq x_3leq x_1right}cup\
left{x_2=0, , , x_1in[0, 1], , , 0 leq x_0leq x_1, , , 0 leq x_3leq x_1right}cup\
left{x_3=0, , , x_1in[0, 1], , , 0 leq x_0leq x_1, , , 0 leq x_2leq x_1right}cup\
left{x_1=x_0, , , x_0in[0,1] , , , 0 leq x_2leq x_0, , , 0 leq x_3leq x_0right}
cup\
left{x_1=x_2, , , x_2in[0,1] , , , 0leq x_0leq x_2, , , 0 leq x_3leq x_2right}cup\
left{x_1=x_3, , , x_3in[0,1] , , , 0 leq x_0leq x_3, , , 0 leq x_2leq x_3right}
$$

Now restricting to one lower dimension simplifies this to the analogous problem
$$
mathcal I=\
left{x_0=0, x_1 in[0, 1], 0 leq x_2leq x_1 right}cup\
left{x_2=0, x_1in[0, 1], 0 leq x_0leq x_1right}cup\
left{x_1=x_0, x_0in[0,1], 0 leq x_2leq x_0 right}cup\
left{x_1=x_2, x_2in[0,1], 0leq x_0leq x_2right}
$$
And substituting the more usual variable names gives
$$
mathcal I=\
left{x=0, y in[0, 1], 0 leq zleq y right}cup\
left{z=0, yin[0, 1], 0 leq xleq yright}cup\
left{y=x, xin[0,1], 0 leq zleq x right}cup\
left{y=z, zin[0,1], 0leq xleq zright}
$$

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 $I$.