Bounded SetDate: 06/24/2003 at 13:57:27 From: Saint Monday Subject: Proof about a bounded set Let S be a set of real numbers. Prove that the following are equivalent: (a) S is bounded, i.e. there exists a number M > 0 such that abs(x) <= M for every x in S. (b) S has an upper bound and lower bound. I need to find the infimum and supremum that implies that there exists an upper and lower bound. This is what I have so far: abs(x) <= M is the same as -M <= x <= M x <= M, upper bound a <= x, lower bound It was suggested to me to find a value for M that would satisfy this, so would M = (abs(x) + 1) work? Date: 06/24/2003 at 14:57:11 From: Doctor Mike Subject: Re: Proof about a bounded set Hi, The infimum is the LARGEST possible lower bound, and the supremum is the SMALLEST possible upper bound, so it is good to use those words only when you mean exactly those specific concepts. To show two statements are "equivalent" means to prove that if you assume one of them is true, the other must be true also. Obviously, such a proof comes in two parts. Namely, Part 1 --- Assume (a) is true, and then prove that (b) is true. This part you did pretty well on before. Assume for notation purposes that the positive "M" is a bound for S. Note, it may NOT be the SMALLEST bound for S. What we are required to prove to show that (b) is true is that S has an upper bound and has a lower bound. You correctly observed that M must be an upper bound, and -M must be a lower bound. Done with this part! Part 2 --- Assume (b) is true, and then prove that (a) is true. For this part, assume for notation purposes that "B" is an upper bound for S, and that "C" is a lower bound for S. That just means that for every x in S, C <= x and x <= B. Now think about this number N. N = max( abs(B) , abs(C) ) That is, N is the larger of the absolute value of B and the absolute value of C. We know that every x in S is <= B so it must also be that x <= abs(B), which in turn is <= N. To summarize, x <= B <= abs(B) <= N Now let's look at what x is greater than or equal to. x >= C is known. We do not know whether C is positive or negative or zero, but we DO know that -abs(C) must be the same as C or smaller than C. So, -abs(C) <= C. Now take a very close look at this LONG thing : x >= C >= -abs(C) >= -max(abs(B),abs(C)) = -N So this x, which can be anything in S, is <= N and >= -N. That's exactly what it means to say that N is a bound for S. Go over this a few times to make sure you understand each step. We are done with this part now. When you are done with both directions (both parts), you are completely done. I hope this clarifies what is going on in this problem. I hope also that you can carry on with this proof as a "pattern" to do similar ones. If you get to a problem where infimum and supremum really are involved, the structure of an "equivalent" type of proof is still the same: two parts. Just the details of the properties to be proved are different. Good luck. Write back again if you want to talk more - Doctor Mike, The Math Forum http://mathforum.org/dr.math/ Date: 06/24/2003 at 15:11:49 From: Saint Monday Subject: Thank you (Proof about about a bounded set) You're a godsend, Dr. Mike. Thank you so much for your time! |
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