Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Well-orderings are rigid proof

  1. Jan 15, 2015 #1
    Hi I've been trying to understand this proof, but there is one step that I don't get at all.

    Proof: Suppose f is an automorphism of (E,<=). Consider a set D, a set of non-fixed points under f. If D is empty, f is an identity mapping. Suppose, toward a contradiction, that D is nonempty. Then D has a least element, say a. Since E is well-ordered, either f(a) < a or a < f(a). Since f(a) < a, f(a) is not an element of D. So f fixes f(a), hence f(f(a)) = f(a). But then f(a) = a since f is injective, contradicting that a is an element of D. The case a < f(a) follows similarly applying the inverse of f.

    Why does f(a) < a imply that f(a) is not a fixed point?
  2. jcsd
  3. Jan 16, 2015 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    To be clear, the book should have said "Consider the case when f(a) < a.".

    Is it possible that f(a) is not a fixed point? If it were not a fixed point, it would be an element of D that is less than a.. But a is defined as the least element of D, so this is impossible.
    Last edited: Jan 16, 2015
  4. Jan 16, 2015 #3
    Thank you very much!. In fact, the book did say "Consider the case when f(a) < a".
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook