Re: Contradiction of bijections as a measure for infinite sets

WM schrieb:

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

Sure - adding a number that is already contained in ℕ, doesn't change
the cardinality of the set, since equal elements are counted as one
element.
But you can do the following described by Cantor:
    "That ℵo is a transfinite number, that is to say, is
     not equal to any finite number μ, follows from the
     simple fact that, if to the aggregate {ν} is added a
     new element e_0, the union-aggregate ({ν}, e_0 ) is
     equivalent to the original aggregate {ν}. For we
     can think of this reciprocally univocal correspondence
     between them: to the element e_0 of the first
     corresponds the element 1 of the second, and to the
     element ν of the first corresponds the element ν + 1 of
     the other. By §3 we thus have
        (2)      ℵo + 1 = ℵo "
(Georg Cantor:
  Contributions to the founding of the theory of tranfinite numbers.
  Dover Publications, 1915; p.104)

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

Infinite sets don't have a "number of elements". This concept (which can
only be used for finite sets) is generalized for infinite sets by the
concept of "transfinite cardinal numbers".
And indeed: there is a bijection from the set of natural numbers ℕ
to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.
    f: ℕ → 𝔼 ,  n ↦ 2n
This function is both injective (or one-to-one) and surjective (or
onto), thus it is bijective.

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.

Rubbish. The cardinality of an infinite set is described by an
transfinite cardinal number and not by a finite "number of elements".
Your problem is: You try to apply facts, that hold for finite sets,
on infinite sets. That doesn't work.
Dieter Heidorn