Hurkyl
Staff Emeritus
Science Advisor
Gold Member
- 14,922
- 28
New theorem, but I think you already anticipated this one.
Suppose f is a bijection between V and V^V.
Then, every function V--->V has a fixed point.
Proof is roughly the same as the previous one I gave:
Let j by a function from V to V.
Define g(x) = f(x)(x)
Let s = f^-1(j o g)
Then, g(s) = f(s)(s) = (j o g)(s) = j(g(s))
Thus, g(s) is a fixed point of j.
Suppose f is a bijection between V and V^V.
Then, every function V--->V has a fixed point.
Proof is roughly the same as the previous one I gave:
Let j by a function from V to V.
Define g(x) = f(x)(x)
Let s = f^-1(j o g)
Then, g(s) = f(s)(s) = (j o g)(s) = j(g(s))
Thus, g(s) is a fixed point of j.