Re: Contradiction of bijections as a measure for infinite sets

WM explained :

Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be added?
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If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both sets have the same number of elements.

Actually, same size set. "Number of elements" is better suited to finite sets.

Then the completion of E resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its elements.

That is not a completion of E. But still the same size set. By your sense of 'complete' the set of even numbers was already 'complete' because no more even numbers could be 'added'.

Then there are more natural numbers than were originally in ℕ.

Nope.