# Help needed. Upper bound of a set proof

• Oct 29th 2012, 01:52 AM
matemauch
Help needed. Upper bound of a set proof
Hello !

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.
• Oct 29th 2012, 05:50 AM
emakarov
Re: Help needed. Upper bound of a set proof
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.
• Oct 29th 2012, 08:27 AM
matemauch
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by emakarov
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.

Let me see if I understood what you said emakarov.
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 !
• Oct 29th 2012, 09:07 AM
emakarov
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by matemauch
Let me see if I understood what you said emakarov.
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 ??

Yes, but you need to understand why, since this is the entire statement that you need to prove. There are two steps. First, since b is the least upper bound, any number c smaller than b is not a upper bound. And the fact that c is not an upper bound means by definition that there exists an element of A greater than it.

Quote:

Originally Posted by matemauch
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.

Given some a in A, how exactly do you define c? And why would c be the supremum of A? And if it is the supremum, maybe it equals b.
• Oct 30th 2012, 04:22 AM
matemauch
Re: Help needed. Upper bound of a set proof
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:
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?
• Oct 30th 2012, 04:51 AM
Plato
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by matemauch
What I need to prove is that for any a in A, a<=b.

No you do not have to show that.
That is the very definition of least upper bound.
If $\displaystyle b=\sup(A)$ then if $\displaystyle x\in A$ it must be true that $\displaystyle x\le b$.
There is nothing to prove, it is a result of the definition.

Now it is true that if $\displaystyle c<b$ it must be true $\displaystyle \exists t\in A$ having the property that $\displaystyle c<t$. Otherwise $\displaystyle b$ would not be the least upper bound.
But we know from the definition that $\displaystyle t\le b$.
• Oct 30th 2012, 05:32 AM
matemauch
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by Plato
No you do not have to show that.
That is the very definition of least upper bound.
If $\displaystyle b=\sup(A)$ then if $\displaystyle x\in A$ it must be true that $\displaystyle x\le b$.
There is nothing to prove, it is a result of the definition.

Now it is true that if $\displaystyle c<b$ it must be true $\displaystyle \exists t\in A$ having the property that $\displaystyle c<t$. Otherwise $\displaystyle b$ would not be the least upper bound.
But we know from the definition that $\displaystyle t\le b$.

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 !
• Oct 30th 2012, 06:02 AM
Plato
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by matemauch
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?

I think that you to post the exact wording of the question.
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 $\displaystyle x\in A$ then $\displaystyle x\le b$.

So please post the exact wording.
• Oct 30th 2012, 07:07 AM
emakarov
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by matemauch
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.

This is not the converse of the original statement. The original statement was (writing ∀ for "for all" and ∃ for "there exists"):

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.
• Oct 30th 2012, 08:26 AM
matemauch
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by emakarov
This is not the converse of the original statement. The original statement was (writing ∀ for "for all" and ∃ for "there exists"):

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.

Thank you very much emakarov. Your explanation was very illustrative !

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.
• Oct 30th 2012, 10:35 AM
matemauch
Re: Help needed. Upper bound of a set proof
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 !
• Oct 30th 2012, 10:37 AM
matemauch
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by emakarov
(∀ 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.

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 !
• Oct 30th 2012, 11:03 AM
emakarov
Re: Help needed. Upper bound of a set proof
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?
• Oct 31st 2012, 06:49 AM
matemauch
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by emakarov
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?

This is what I came up with:

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.
• Oct 31st 2012, 10:17 AM
emakarov
Re: Help needed. Upper bound of a set proof
Quote:

Originally Posted by matemauch
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.

This works. Instead of funding some c between u and b, you could also just put c = u in the assumption and get an a ∈ A such that a > c = u.