Consider the set of two elements P₀={ a b } where a and b, b>a, are reals, that is, a,b∊ℝ.
We can assign the cardinals 1 and 2 to a and b respectively. That is, we can count the elements of P₀ and C(b)=n(P₀), where C denotes cardinality.
We can expand the set P by adding c where c=(a+b)/2, making P₁={ a c b } with cardinalities 1, 2 and 3. C(b)=n(P₁)=3. Then we create P₂={ a d c e b } where d=(a+c)/2 and e=(b+c)/2. C(b)=n(P₂)=5.
If we repeat the process we end up with C(b)=n(Pᵪ)=2ᵡ+1. For example, if X=10, C(b)=n(P₁₀)=1025.
There are an infinite number of reals, as X→∞, and there will be a corresponding infinite number of cardinals.
Between a and b there will be an infinite number of reals (e.g., a=-1, b=1, c=0, d=-0.5, e=0.5, f=-0.75, g=-0.25, h=0.25, i=0.75, etc.).
There are an infinite number of natural numbers, but the set of reals includes the set of natural numbers, making the set of natural numbers a subset of the reals. The “infinities” of the natural numbers and the reals are not the same. Also, there is no first element a (-∞) or last element b (∞), so it would not be possible to assign cardinality to the elements in an infinite set. I think cardinality can only be applied to a finite subset of reals and a corresponding finite set of cardinal natural numbers.
Another argument is that the natural numbers are strictly positive, but the range of reals extends from -∞ to +∞, so that even if the set of cardinals were to be sufficient to count the reals from 1 to ∞ we would need the same set again to count the reals from -∞ to 0. Therefore there can be no set of cardinals mapping to the reals.
