Consider the set given by $\{(x, y, z) \in\Bbb R : |x| + |y| + |z| \le 1\}$. Show that it is a polyhedron. Count the number of faces of dimensions $0, 1, 2$ (That is, count (separately) the number of vertices, edges and facets).
2026-03-27 03:44:18.1774583058
Consider the set given by $\{(x, y, z) \in\Bbb R : |x| + |y| + |z| \le 1\}$. Show that it is a polyhedron.
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The set is bounded by $8$ planes (assume signs for $x,y,z$ then remove the absolute values, you get $8$ equations $\pm x\pm y\pm z=1$ for the boundaries. These define planes.
Now, for each of these planes, you can find $3$ points in the set $\{(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)\}$ which belong to the plane. Which polyhedron does that yield?