Let us show that S⊂C. If P(x,y,z)∈S, then x2+y2+z2<1. Since x2≥0 and y2+z2≥0, the inequality x2+y2+z2<1 implies x2<1. It follows that −1<x<1. By analogy, y2≥0, x2+z2≥0 and x2+y2+z2<1 imply y2<1, and hence −1<y<1. In the same way, −1<z<1. We conclude that P(x,y,z)∈C, and hence S⊂C.
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