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Math Help - Dedekind Cuts?

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    Dedekind Cuts?

    Given a real number a, define S≡{x|x in Q, x < a}. Prove that a = sup S using dedekind cuts.

    Here is a summary of the proof I used:

    If ∃ B|B is an upper bound of S and B < a
    then (B+a)/2 is not in S.
    But (B+a)/2 < a, so we have proof by contradiction.

    But I don't know how dedekind cuts would be used in this, since I don't really understand what they are and they aren't used in my text book.

    If anyone could point me in the right direction, I would appreciate it.
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    You have asked a very complicated question. The complication results from the wide varieties of notation is use when discussing Dedekind Cuts, Your best bet is find a good text in your library that uses Cuts. A presently popular text is Elementry Analysis by Ken Ross it has a very brief discussion. A very full development can be found in John Randolph’s Basic Real and Abstract Analysis. It has some 15 pages on Cuts but it is an older text.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sjhlax89 View Post
    Given a real number a, define S≡{x|x in Q, x < a}. Prove that a = sup S using dedekind cuts.

    Here is a summary of the proof I used:

    If ∃ B|B is an upper bound of S and B < a
    then (B+a)/2 is not in S.
    But (B+a)/2 < a, so we have proof by contradiction.

    But I don't know how dedekind cuts would be used in this, since I don't really understand what they are and they aren't used in my text book.

    If anyone could point me in the right direction, I would appreciate it.
    Plato is right. (I used Ken Ross' book, it doesn't have much on Dedikind cuts and I never did that in my class either, since it's a stared section and we skipped all of those). I do recall that one of the properties of a Dedikind cut though, is that if a set S is a Dedikind cut and we have s \in S, s \in \mathbb{Q}. If we have r \in \mathbb{Q} such that r < s then r \in S (or something like that, I don't have my book with me.

    If I am correct though, we can do a proof by contradiction using the Denseness of \mathbb{Q} property.

    Assume to the contrary that a \ne \sup S, but rather, some other element x = \sup S. Then, by the definition of the supremum, we must have that x < a...

    Now apply the Denseness of \mathbb{Q} property to arrive at a contradiction. Can you continue?
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