When asked to prove your if statement, your "proof" was to simply say it exists but somehow that was not an assumption? That is neither an application of math or logic, so you are not really employing either one in this conversation. You either prove your statements or you assume them.
You might be surprised at how many assumptions we make just to do math.
Sorry, I was assuming you were aware of the basics. Yes, I am quite aware of the assumptions required to do math.
OK, in Zormelo Fraenkl Set Theory with Choice (ZFC), which is the standard set of axioms for mathematics, there are two relevant axioms:
1. If A is a set, there is another set that contains all subsets of A. We call that set the power set of A and write it as P(A).
2. There is a set that has the empty set as an element and whenever x is an element, then x u {x} is an element.
The set of natural numbers, N, is defined to be the intersection of ALL sets satisfying the second property.
Of course, you have to define what it means to be infinite (and there are several different possible definitions for this, all equivalent in ZFC). So, we can say that a set A is infinite if there is a one-to one function f:A->A which is not onto.
With this definition, the function f:N->N defined by f(x) = x u {x} can be shown to be one-to-one and not onto, showing that N is infinite.
It is also true that P(N) is infinite by a similar trick.
So, another way of stating the continuum hypotheses is that every infinite subset of P(N) is either in one-to-one correspondence with N or with P(N).
If you really need me to start at the axioms and deduce all of this, I can, but it is probably best for another thread.
