Re: Contradiction of bijections as a measure for infinite sets

Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both sets have the same number of elements. Then the completion of E resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its elements. Then there are more natural numbers than were originally in ℕ.
Regards, WM