Well-orderings are rigid proof

  • Thread starter wj2cho
  • Start date
  • #1
20
0

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
7,020
1,244
Since f(a) < a, f(a) is not an element of D.
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:
  • #3
20
0
Thank you very much!. In fact, the book did say "Consider the case when f(a) < a".
 

Related Threads on Well-orderings are rigid proof

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
12
Views
4K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
2
Views
5K
Replies
2
Views
765
Replies
2
Views
3K
Top