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Cardinality != Density?
The theory: For the infinite set of natural numbers Aleph-naught indicates it's cardinality, and therefor any other set that is countable (using natural numbers) must then also have the same cardinality, which essentially means they are the same size in relation to infinity.
Take Galileo's paradox:
the set of squares is a subset of the set of natural numbers. Find a set that contains both as subsets (in this case it would be the same as the set of natural numbers):
A = { 1,2,3,[4],5,6,7,8,[9],10,11,12,13,14,15,[16],17,... } (the square brackets just indicate the square values)
Counting the elements of each of the natural numbers and the squares in the sequence in which they occur in set A with separate counters to the sequence counter for set A, will give after 10 elements of A was counted, 10 elements of the (sub)set of natural numbers, and 3 elements of the subset of squares, and after 17 elements of A the values would be 17 and 4 for natural numbers and squares respectively.
From these counts a rate-of-increase (r-o-i) for the set of natural numbers and the set of squares can be calculated: n = r-o-i(N after 17 elements) = 1 s = r-o-i(Squares after 17 elements) = 4/17
if n can be shown to be consistently greater than s after any arbitrary count of elements in A greater than 1, then a valid conclusion would be that at any count of elements of A greater than 1, that n would be larger than s, and therefor that the cardinality of the set of natural numbers would be larger than that of the set of squares, even at infinity.
The only problem of comparing the cardinality of any two or more countable sets, would be to find such countable set A of which these sets are subsets and then to calculate their respective rates-of-increase.
Would the foregoing be a valid/reasonable approach to compare the relative sizes of countable infinite sets?
I was pointed to this earlier question, which seems my question is a dupe of..