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Math Help - Help needed. Upper bound of a set proof

  1. #1
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    Question 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.
    Thank you in advance !!
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    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.
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by emakarov View Post
    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 !
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by matemauch View Post
    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 View Post
    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.
    Thanks from matemauch
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    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?
    Thank you in advance !
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by matemauch View Post
    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 b=\sup(A) then if x\in A it must be true that x\le b.
    There is nothing to prove, it is a result of the definition.

    Now it is true that if c<b it must be true \exists t\in A having the property that c<t. Otherwise b would not be the least upper bound.
    But we know from the definition that t\le b.
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    Re: Help needed. Upper bound of a set proof

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

    Now it is true that if c<b it must be true \exists t\in A having the property that c<t. Otherwise b would not be the least upper bound.
    But we know from the definition that 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 !
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by matemauch View Post
    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 x\in A then x\le b.

    So please post the exact wording.
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by matemauch View Post
    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.
    Thanks from matemauch
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by emakarov View Post
    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.
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    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 !
    Last edited by matemauch; October 30th 2012 at 10:37 AM.
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by emakarov View Post
    (∀ 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 !
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    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?
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by emakarov View Post
    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.
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    Re: Help needed. Upper bound of a set proof

    Quote Originally Posted by matemauch View Post
    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.
    Thanks from matemauch
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