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Thread: Extensionality and well-founded

  1. #1
    Super Member Showcase_22's Avatar
    Sep 2006
    The raggedy edge.

    Extensionality and well-founded

    Let \vartriangleleft and \hat{\vartriangleleft} be relations on sets X and \hat{X} respectively.

    The structure (X, \vartriangleleft) is:

    1). extensional if \forall x \in X \forall y \in X [\forall z \in X(z \vartriangleleft x \leftrightarrow z \vartriangle y) \rightarrow x=y],

    2).well-founded if \forall A[\phi \neq A \subset X \rightarrow \exists a \in A \forall x \in A  (x \vartriangleleft a)]

    Show that if (X, \vartriangleleft) and (\hat{X}, \hat{\vartriangleleft}) are extensional and well-founded, then there is at most one isomorphism between them.

    Hint: Consider \{ x \in X | f(x) \neq x \} for an automorphism f of (X, \vartriangleleft).

    I have absolutely no idea how to do this question. I can't see how the hint is helpful. If I consider the automorphism, i'm not considering (\hat{X}, \hat{\vartriangleleft}) which I need to do.

    Can anyone point me in the right direction?
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  2. #2
    Senior Member
    Nov 2010
    Staten Island, NY
    Well an isomorphism has an inverse. So if you have two isomorphisms g and h, you can consider the automorphism h^{-1}\dot g.

    If the two automorphisms are distinct, then the composition I suggested is an automorphism which is not the identity. So it suffices to show that the only automorphism of X is the identity map. Use the hint to show this.
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