There is no need to consider two cases. If c < b, then c is not an upper bound since b is the least upper bound. Therefore, there exists an a in A such that c < a.
I am trying to prove the following 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.
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 help will be very much appreciated.
Thank you in advance !!
By hypothesis b is the supremum of A. This guarantees that for any given c<b there is an a in A such that c<a ??
I also found out that I can prove the statement by contradiction letting c>a for any a in A. The result would be that c is the supremum of A which is a contradiction since, by hypothesis b is the supremum of the set A.
Your comment will be valuable. Thank you !
I think I understand your statement. Thank you !
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:
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 !
That is the very definition of least upper bound.
If then if it must be true that .
There is nothing to prove, it is a result of the definition.
Now it is true that if it must be true having the property that . Otherwise would not be the least upper bound.
But we know from the definition that .
I think I am a little confused here.
As far as I understand I need to prove that b is the supremum of A. In the givens the only information I have of b is that it is a real number. So as far as I understand I need to do something in order to show that for all a in A, a<=b. Isnīt that right? Otherwise how can I guarantee that a is not bigger than b?
Thank you for the help !
In the OP you wrote "If b is the supremum (least upper bound) of the set A..."
That means the you are given that b is the supremum of the set A.
You are therefore given that if then .
So please post the exact wording.
b = sup A => ∀ c < b ∃ a ∈ A. c < a (*)
In fact, the assumption b = sup A is too strong; it is enough to assume half of the definition of sup, namely, that b is <= all upper bounds of A. That is, we don't have to assume that b is itself an upper bound.
The converse of (*) is the following.
(∀ c < b ∃ a ∈ A. c < a) => b = sup A (**)
Again, (**) is not provable because the conclusion is too strong. From the given assumption we can only prove that b is <= all upper bounds of A. Either the conclusion has to be changed as in the previous sentence, or the fact that b is an upper bound of A has to be added to the assumption.
Now, the important point is that your version of (**) is this.
∀c [(c < b and (∃ a ∈ A. c < a)) => b = sup A] (***)
An essential difference between (**) and (***) is that in (**), the scope of ∀ c is limited to the assumption, while in (***) it extends to the end of the formula. This is what happens when you say "Let c be real" in the beginning. As a result, in (***) you need to show b = sup A for each single c. In contrast, the assumption in (**) requires that something holds for all c. This makes the assumption in (**) much stronger. And, of course, (***) is not a converse of (*) because in (*) the scope of ∀ c is limited to the conclusion only.
Plato: You are right. In the OP I wrote "If b is the supremum (least upper bound) of the set A..." and I got some help in proving that statement. It was later on that I wrote "Now I am trying to prove the statement in the other direction..." It was my intention to be write clear since the beginning. Sorry if I made you be confused.
If we change the conclusion as you suggested (to prove b is <= all upper bounds), I am still finding difficulties proving it.
Could you please give me a hint ?
Thank you very much for your help !
We need to prove
(∀ c < b ∃ a ∈ A. c < a) => ∀ u (u is an upper bound of A => b <= u)
So assume ∀ c < b ∃ a ∈ A. c < a, fix an arbitrary u and assume that u is an upper bound of A, i.e., ∀ x ∈ A, x <= u. We need to show that b <= u. Suppose the contrary, i.e., u < b. How can we use the assumption to derive a contradiction?
by assumption b>u which means there is a c such that u<c<b. By the givens there is an a such that c<a and therefore u<a, which is a contradiction since u is an upper bound of the set A i.e. for all a in A a<=u.
Therefore b>= u.
What do you think?
Thank you for your help ! I have been learning a lot during this exchange of posts.