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1) A = { x[itex]\in[/itex] ℝ: (x-a)(x-b)(x-c) < 0 } , a<b<c. Find the supA and infA.

In the solution of his example he says. It is easy to see that A = (-∞,a)[itex]\cup[/itex](b,c) so infA = -∞ and supA = c. How did he find that A = (-∞,a)[itex]\cup[/itex](b,c) ?

2) B = {1 +(-1)

^{n}:n[itex]\in[/itex] N}. Find the supB and infB.

In the solution he says. Obviously B={0,1} which we can compute if we make n=2k (even) and n=2k+1 (odd)

so infB=0 and supB=1. So if n=2k wouldn't B be 2? Is this a mistake my professor made?