# Cardinal Exponentiation

• May 1st 2010, 03:22 AM
Fibon
Cardinal Exponentiation
Hello,

I was wondering if someone could help me show the following:

kappa, lambda and mu (for brevity denoted k,l and m) are 3 cardinals, show that:

1) k^(l+m)=k^l k^m
2) (k l)^m=k^m l^m

I have given the matter some thought already, but I feel highly insecure about it, and I was therefore wondering if someone could please look at it, and give me some help if it is wrong:

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For (1)

Let k=card(A), l=card(B) and m=card(C). Then to show (1) we have to show that A^(B disjoint union C) ~ A^B x A^C

By definition in my notes, B disjoint union C = ({0}xB)U({1}xC)

Now

A^(B disjoint union C) ={f:f is a function from ({0}xB)U({1}xC) to A}
A^B = {g:g is a function from B to A}
A^C = {h:h is a function from C to A}

The map

A^(B disjoint union C) --> A^B x A^C
((0,b)->a) -> (b->a,c->a)
((0,c)->a) -> (b->a,c->a)

is a bijection (the inverse is just the reverse). Thus A^(B disjoint union C) ~ A^B x A^C
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For (2):

In this case we want to show that (AxB)^C ~ A^C x B^C

Again we have

(AxB)^C = {f: f is a function from C to A x B}
A^C = {g: g is a function from C to A}
B^C = {h: h is a function from C to B}

The map

(AxB)^C --> A^C x B^C
(c ->(a,b)) -> ((c->a),(c->b))

is a bijection (the inverse is just the reverse map). Thus (AxB)^C ~ A^C x B^C

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