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Math Help - Least upper bounds of subsets

  1. #1
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    Least upper bounds of subsets

    I'm supposed to prove that if A is a subset of B then sup(A) <or= to sup(B), provided that these least upper bounds exist.

    So I started by using the definition of a least upper bound. I let x = sup(A) therefore by definition there exists a x in B such that for ever a in A, a <or= to x.
    also there exists a c in B such that for every a in A, a <or= c and c >or= x.

    My problem is though I'm not completely sure though whether c is necessarily in B. because then I let y=sup(B) and say because c in B, by definition y >or= c.

    Then by transitivity x <or= y.

    Also if there are any ideas of how this proof could be done better I'd appreciate it because this doesnt seem solid enough.
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  2. #2
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    actually I could shorten it even more and say because x in B then by definition of y=sup(B) for every x in B y >or= x.

    But then again I'm not sure I can say x is in B.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nataliemarie View Post
    I'm supposed to prove that if A is a subset of B then sup(A) <or= to sup(B), provided that these least upper bounds exist.

    So I started by using the definition of a least upper bound. I let x = sup(A) therefore by definition there exists a x in B such that for ever a in A, a <or= to x.
    also there exists a c in B such that for every a in A, a <or= c and c >or= x.

    My problem is though I'm not completely sure though whether c is necessarily in B. because then I let y=sup(B) and say because c in B, by definition y >or= c.

    Then by transitivity x <or= y.

    Also if there are any ideas of how this proof could be done better I'd appreciate it because this doesnt seem solid enough.
    Problem: Let A\subset B\subset\mathbb{R}. Prove that \sup \text{ }A\le\sup\text{ }B

    Proof: By definition \sup\text{ }B is an upper bound for B, therefore \forall x\in B\quad x\le \sup\text{ }B. Also, by definition A\subset B\Longleftrightarrow \bigg[\left(x\in A\right)\implies \left(x\in B\right)\bigg]. Therefore \sup\text{ }B is an upper bound for A. Thus, by definition, \sup\text{ }A\le\sup\text{ }B
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