>>26
25 here, everything you just said was wrong.
(quote) Wow. First of all, the set-theoretical definition of the natural numbers identifies 0 as { {} }, not {}. 1 is identified with { {}, { {} } }.
Answer: Firstly, 0 is defined as {} not {{}}. The point is that "0" (the cardinal number) is equivalent (ie. has the same "number of elements", intuitively speaking) as a set that we can say has 0 elements. So if what you said was true, the set {x}~{{}} (because the bijection x ~> {} allows this equivalence) so #{x}=0, which is nonsense. Look this up anywhere and it will tell you that 0 is defined as {}.
>(quote) Your definition of exponentiation would give 0^1 = #{all functions from 0 to 1} = 2.
Secondly, it is not _MY_ defintion of exponentiation: it is one the THE standard formal definition of exponentiation for natural numbers (and cardinal numbers in general).
I will clarify it. In this definition x^y = #{all functions from x to y}. Hence 0^1 = #{all functions from {{}} to {}} = #{} = 0 (there are no functions from {{}} to {}). In general, in fact, by this definition 0^y = 0 where y>0. As I showed before, when y=0, 0^y=1.
You swapped the variables (which is my fault for not clarifying the definition), so what you should have got is 1^0 = #{all functions from {} to {{}}.} = #{ the only one is the empty function, the function with domain and range = {}.} = 1.
You didn't get this answer because you had 0={{}} and 1={{}, {{}}.} but 1={{}} and 2={{}, {{}}.}. So what you actually were trying to work out is 2^1=2, which is right!!!