Re: Contradiction of bijections as a measure for infinite sets

Am 24.03.2024 um 21:11 schrieb WM:

Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be added?

Well, IN is defined this way. I mean, as the set of _all_ natural numbers.

If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both sets have the same number of elements.

Of course. #IN = aleph_0 = #E.

Then the completion of E resulting in E* = {1, 2, 3, 4, 5, 6, ...}[.] [Hence the number of E*'s elements] would [be twice] the number of [E's] elements.

Exactly!
Hint: The number of elements in E is aleph_0. And 2 * aleph_0 = aleph_0 + aleph_0 = aleph_0.

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

Nope, since 2 * aleph_0 = aleph_0.
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