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Thread: WellOrdering Theorem Proof

  1. #1
    Mar 2013

    WellOrdering Theorem Proof

    I've been going through a proof of Zermelo's Theorem of Well Ordering, but there's a step that I don't quite understand:
    WellOrdering Theorem Proof-proofsnippet.png
    It's showing that \phi is well-ordered. Why does it follow from the fact that T is in V? From this I gather that every non-empty subset has a largest element. I can see that this defines a well-ordering, but not that the inclusion order we're working with is one. I thought maybe they meant a well-ordering exist, but at the end they construct an order on S that is preserved under inclusion in \phi, so they must mean that inclusion is the well-order.

    Little help clarifying, please?
    Attached Thumbnails Attached Thumbnails WellOrdering Theorem Proof-proofsnippet.png  
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  2. #2
    MHF Contributor
    Dec 2012
    Athens, OH, USA

    Re: WellOrdering Theorem Proof

    A total ordering < on a set is a well ordering iff for any non-empty subset V, there is a least element of V. That is there is v in V with v\leq w for all w in V. In particular for set inclusion as <, to show V has a least element, it is clear that T= intersection of all elements of V (subsets of the original S) has the property that T\leq w for all w in V. So the only thing to verify is that T is in V.
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