Re: set question

On Mon, 29 Jan 2007, Dave L. Renfro wrote:
On Jan 29, 4:39 pm, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:

The empty set is a bijection from the empty set to the
empty set. Remember what it means for f to be a function
from set A to set B -- f is a subset of A x B that has
a certain property. The empty set is a subset of
(empty set) x (empty set), and "empty set" is vacuously
a bijective function. (The definitions of "function" and
"inverse is a function" are certain conditional statements
that, in this case, are vacuously true.)

And now we have a proof that 0^0 = 1 for cardinal numbers.
(Well, at least a proof that 0^0 > 0.)

As nulset x nulset = nulset and
A = { f:nulset -> nulset } subset P(nulset x nulset)
A = { nulset }

I think this is going to do it for me today. It's time to go home. (And
I doubt I'll be getting on the internet this evening.)

No problem, this will wait until tomorrow.

Assume K nonnul. nulset x K = nulset = K x nulset
B = { f:K -> nulset } = nulset
C = { f:nulset -> K } = { nulset }

From A,B and C we have
0^0 = 1
0^|K| = 0
|K|^0 = 1
.

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