“summation of {i = 0} to n of (n combination i) = 2^n”
This is not a proof that “2^{aleph_0}” is the cardinality of the set of the subsets of natural numbers. You assume it works in the infinite cardinal case (without proving it), and then say that you thus proved it. You got confused by notation.
“I shall proffer a mathematical proof to show that for any infinite set of cardinality aleph_0 (the cardinality of the set of natural numbers) there are aleph_1 (2aleph_0) distinct infinite subsets.”
No. 2^{aleph_0} is /by definition/ the cardinality of the set of the subsets of the natural numbers.
It’s named that way to allow the intuition of “summation of {i = 0} to n of (n combination i) = 2n” to work with cardinalities.
“aleph_1 (2^{aleph_0})”
aleph_1 = 2^{aleph_0} has been shown to be independent from ZFC. ie, if we haven’t worked within inconsistent math for that past 60 years, what you just said is unprovable.
You might have confused aleph and beth numbers.
“summation of {i = 0} to n of (n combination i) = 2^n”
This is not a proof that “2^{aleph_0}” is the cardinality of the set of the subsets of natural numbers. You assume it works in the infinite cardinal case (without proving it), and then say that you thus proved it. You got confused by notation.
“I shall proffer a mathematical proof to show that for any infinite set of cardinality aleph_0 (the cardinality of the set of natural numbers) there are aleph_1 (2aleph_0) distinct infinite subsets.”
No. 2^{aleph_0} is /by definition/ the cardinality of the set of the subsets of the natural numbers. It’s named that way to allow the intuition of “summation of {i = 0} to n of (n combination i) = 2n” to work with cardinalities.
“aleph_1 (2^{aleph_0})”
aleph_1 = 2^{aleph_0} has been shown to be independent from ZFC. ie, if we haven’t worked within inconsistent math for that past 60 years, what you just said is unprovable. You might have confused aleph and beth numbers.
I have edited the article to be more accurate.