pasmith said:
You don't need to use a proof by contradiction.
Show that there exists a bijection from the power set of [itex]K[/itex] to the set of functions from [itex]K[/itex] to [itex]\{0,1\}[/itex]. This set is a proper subset of [itex]F^{*}[/itex], since the codomain of a function in [itex]F^{*}[/itex] can be any non-empty set.
Why does this imply that there is no bijection from [itex]K[/itex] to [itex]F^{*}[/itex]?
Ok, I'll try, I was not able to find an bijection from P(K) to the functions, but an injection, but I think it still works.
Lemma: card (P(K)) <= card L
L is the set of functions with codomain {0,1}, and domain K.
Define F2: P(K)-> L
F2(K') = {(i,j) : i is an element of K: j = 1 if i is an element of K', else j = 0}
K' is a subset of K.
F2 gives a function which has domain K, and this new function is 1 only if it is evaluated in K'.
it is must be invertible because if F2(K')=F2(K''), then K'=K''.Now we have that:
Card K < Card(P(K)) <= card(L) <= card(F*), since L is a subset of F*.
We can't have a bijection from K to F*, because then we could create an injection from P(K) to K, which is impossible.
Out of curiosity, how would you make the F2 function also surjective?
EDIT:
Haha, I see that the function I defined is actually surjective. Thanks for the help, it was a very smart trick!