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Math Help - Supremum

  1. #1
    Member billa's Avatar
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    Supremum

    Let A,B be subsets of an ordered field F
    Let A+B denote the set {(x+y)|x is in A, y is in B)}
    Let q=sup A, p=sup B
    Show that supA+B = q+p

    If this were a test question (which it isn't!), and I turned the following in, what grade do you think would I receive and why?.


    Here is what I need to show:
    (i) :  (\forall x \in A+B) x \le (q+p) - this says that (q+p) is an upper bound
    (ii) :  (\forall t < (q+p))(\exists x \in A+B) x>t - this says that nothing less than (q+p) is an upper bound

    Here is the proof:
     <br />
(\forall x \in A) p \ge x<br />
( by definition )
    <br />
(\forall y \in B) q \ge y    <br />
( by definition )

    <br />
\implies (\forall x \in A, y \in B) p+q \ge x+y  \implies (\forall (x+y) \in A+B) p+q \ge (x+y)<br />

    (so (i) is true)

    <br />
(\forall t1 < p)(\exists x \in A) x > t1<br />
(By definition)
    <br />
(\forall t2 < q)(\exists y \in B) y > t2<br />
(By definition)
    <br />
\implies (\forall t=t1+t2 < p+q) (\exists x \in A, y\in B) x+y > t1+t2=t<br />
( note: t1 < p t2<q )
    <br />
\implies (\exists (x+y) \in A+B) (x+y) > t<br />

    So (ii) is true
    Therefore sup A+B = q+p


    Thanks :-)
    I am trying to go thru a book on my own before school starts, but I am always worried because there is no answer key.
    Last edited by billa; July 21st 2009 at 11:40 PM. Reason: Tried to make more followable after the 4th post - and again after the 5th
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  2. #2
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    Quote Originally Posted by billa View Post
    Let A,B be subsets of an ordered field F
    Let A+B denote the set {(x+y)|x is in A, y is in B)}
    Let q=sup A, p=sup B
    Show that supA+B = q+p
    It should be clear to you that \sup \left( {A + B} \right) \leqslant q + p.

    So suppose that \sup \left( {A + B} \right) < q + p.
    Then let \varepsilon  = \frac{{q + p - \sup \left( {A + B} \right)}}{2}.
    Then
    \left( {\exists a' \in A} \right)\left[ {q - \varepsilon  < a' \leqslant q} \right]\;\& \,\left( {\exists b' \in B} \right)\left[ {p - \varepsilon  < b' \leqslant p} \right]

    Now there is a contradiction lurking in all that.
    Can you find it?
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  3. #3
    Member billa's Avatar
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    That was a cool and short way of doing it

    When you add the two inequalities you get sup A+B < sup A+B

    BUT, since you demonstrated that you clearly understand the problem, can you read my proof and tell me if it is right. I had no doubt that what i was trying to prove was actually true, but I am not so sure that my proof is right.

    Thanks
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  4. #4
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    Quote Originally Posted by billa View Post
    BUT, since you demonstrated that you clearly understand the problem, can you read my proof and tell me if it is right. I had no doubt that what i was trying to prove was actually true, but I am not so sure that my proof is right.
    Actually no. I fear that I connot follow what you have done.
    Maybe I am blinded by what I know about this problem.
    Sorry.
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  5. #5
    Super Member Gamma's Avatar
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    question about the field

    Don't you need some sort of assumption on the completeness of the field to be sure this exists? Or at least that it is a linear continuum or something?

    I think your proof looks okay provided your field is nice enough. you have the inequalities backwards in those two lines under (so (i) is true). Also I think you need to be a little more careful with your proof writing. I am assuming t1\in A and t2\in B. When you say let t=t1+t2< p+q you are only sure that at least one of t1 or t2 is less than p or q respectively. It could be the case that t1=p and t2<q. But the main idea of the proof looks okay to me.

    I am just slightly concerned about the lack of assumptions on the field itself, like how do you know for sure that if you have t_1<p for t_1,p \in F that there exists some f\in F such that t_1 < f < p?
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  6. #6
    Member billa's Avatar
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    Thanks for the reply :-)
    I see what you mean about being more clear on t1 and t2, and I cannot believe i put the inequalities backwards - I read over that post many times lol

    Quote Originally Posted by Gamma View Post
    Don't you need some sort of assumption on the completeness of the field to be sure this exists? Or at least that it is a linear continuum or something?

    I am just slightly concerned about the lack of assumptions on the field itself, like how do you know for sure that if you have t_1<p for t_1,p \in F that there exists some f\in F such that t_1 < f < p?
    I don't think so, here is why (warning:this could be quite wrong because I am very tired)
    First, t_1 < f  \le p and f \in A
    If f didn't exist, then t1 would be an upper bound on A, and p wouldn't be the least upper bound
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  7. #7
    Super Member Gamma's Avatar
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    p could be in the set A too though, so t_1 is not an upper bound either. for instance sup[0,1]=1\in [0,1].

    I am just kind of worried about like what if there were only two elements in A or something, you might not be able to find an element between the sup and your t1.

    I am not sure exactly if there are ordered fields out there like that, all finite fields in the algebraic sense are of the form \mathbb{Z}_{p^n} for a prime p in terms of isomorphism class, but I have no clue what kinds of order topologies could be put on them. I am not sure how one would put a proper ordering on fields like that where something like that could happen.

    I am just thinking the field needs to be a linear continuum at least for these two proofs that have been supplied to go through. I am pretty sure the statement has to still be correct; however I am not sure how to prove it without sort of constructing a new element that I cannot be sure necessarily exists.
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  8. #8
    Member billa's Avatar
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    Quote Originally Posted by Gamma View Post
    p could be in the set A too though, so t_1 is not an upper bound either. for instance sup[0,1]=1\in [0,1].

    I am just kind of worried about like what if there were only two elements in A or something, you might not be able to find an element between the sup and your t1.
    If p \in A, then can't f = p ?

    I think that if you can't find an x \in A between t1 and sup A then sup A isn't the least upper bound.

    In the set {0,1} in an ordered Field, I think you can always pick 1 for x, and no matter what your value of p>t1 \implies 1=x>t1


    I am not sure exactly if there are ordered fields out there like that, all finite fields in the algebraic sense are of the form for a prime p in terms of isomorphism class, but I have no clue what kinds of order topologies could be put on them. I am not sure how one would put a proper ordering on fields like that where something like that could happen.
    I don't know what any of these things are
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  9. #9
    Super Member Gamma's Avatar
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    proof

    I think it may be best to do this one by contradiction. Try this on for size.

    Say suppose there exists s=s_1+s_2 \in A+B which is an upper bound and s=s_1 + s_2<p+q\Rightarrow (p-s_1)+(q-s_2)>0 which means at least one of the things in parentheses is positive so without loss of generality suppose 0<p-s_1 or in other words, s_1<p and s_1 \in A. But then by your definition of LUB, there exists s_1'\in A such that s_1<s_1'.

    Now we consider the point s_1' + s_2 \in A +B and see that s_1'+s_2>s_1+s_2=s contradicting the fact that s was in fact an upper bound for A+B.
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