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

    The structure $\displaystyle (X, \vartriangleleft)$ is:

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

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

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

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

    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 $\displaystyle (\hat{X}, \hat{\vartriangleleft})$ 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 $\displaystyle g$ and $\displaystyle h$, you can consider the automorphism $\displaystyle h^{-1}\dot g$.

    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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