How to Prove an Upper Bound for a Set of Real Numbers?

matematiker
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Homework Statement


Let A be a set of real numbers. If b is the supremum (least upper bound) of the set A then whenever c<b there exist an a in A such that a>c.


Homework Equations





The Attempt at a Solution



I considered two cases. The first one when the supremum b is attained by the set A. In this case there exists an a belonging to A such that a=b and the statement is proved.

In the second case the supremum is not attained by the set A, so for all a that belong to A, a<b. Here is where I get stucked. I cannot come up with an idea of an a larger than c but smaller than b.

Any hint in the right direction will be very much appreciated. Thank you !
 
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How about a proof by contradiction? If for c≥a for all points a in A, what does that tell you about the least upper bound?
 
clamtrox said:
How about a proof by contradiction? If for c≥a for all points a in A, what does that tell you about the least upper bound?
That would mean b is not the least upper bound since c is smaller than b and greater or equal to any a in A which is a contradiction since by hypothesis b is the supremum of A.

Awesome !
Thank you clamtrox !
 
matematiker said:
That would mean b is not the least upper bound since c is smaller than b and greater or equal to any a in A which is a contradiction since by hypothesis b is the supremum of A.

Awesome !
Thank you clamtrox !



Now I am trying to prove the statement in the other direction:
Let a,b,c be reals and let A be a set of real numbers. If c<b and there is an a in A such that a>c then b is the supremum of A.

From the givens I know that:
c<b
There is an a in A such that a>c.

What I need to prove is that for any a in A, a<=b.
I took a>c from the givens but that is where I get stucked because I do not know how to show that this a is greater or equal to b.

Do you have any hint?
Thank you in advance !
 
matematiker said:
Let a,b,c be reals and let A be a set of real numbers. If c<b and there is an a in A such that a>c then b is the supremum of A.

There's something wrong here... Let a=1, A={1}, c=0 and b=42. Then c<b and a>c exists, yet b is not the supremum. Maybe it should say "If for all c<b there exists a in A s.t. a>c ... "
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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