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Bounded Set

Date: 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 

(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


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 

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!
Associated Topics:
College Definitions
High School Definitions
High School Sets

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