Re: Question?

Zaljohar@xxxxxxxxx writes:

Hi all,

Let B={1,2}
Let N={1,2,3,....}
Let C={ f | f:N->B }

What is the proof in ZF that |C|>|N| ?

We could prove this by first showing that |C| = |P(N)|, but we can
also show it directly.

I assume you agree that |C| >= |N|. We'll show that there is no
surjection from N to C.

Let g:N -> C be given. Let R = { n in N | g(n)(n) = 1 }. Let
h: N -> B be the function


h(n) = 1 if g(n)(n) = 2
2 if g(n)(n) = 1

Then h is an element of C. We claim that there is no n such that
g(n) = h.

Let n in N. If g(n) = h, then g(n)(n) = h(n), but h(n) = 1 iff
g(n)(n) != 1.
--
Jesse F. Hughes
"How come there's still apes running around loose and there are
humans? Why did some of them decide to evolve and some did not? Did
they choose to stay as a monkey or what?" -Kans. Board of Ed member
.