Mueckenheim published in sci.math three proofs of dark numbers. For convenience I simply copy them here. Are they valid?
(1) Cantor has proved that all positive fractions m/n can be enumerated by all natural numbers k:
k = (m + n - 1)(m + n - 2)/2 + m. (*)
This is tantamount to enumerating the positive fractions by the integer fractions of the first column of the matrix
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
...
Of course also the integer fractions belong to the fractions to be enumerated. Therefore his approach is tantamount to exchanging X's and O's in the matrix until all O's have disappeared:
X, O, O, O, ...
X, O, O, O, ...
X, O, O, O, ...
X, O, O, O, ...
...
In fact by application of (*) all O's are removed from all visible or definable matrix positions. However it is clear that, by simple exchanging O's with X's, never an O will be removed from the matrix. This shows that the O's move to invisible, i.e., undefinable matrix positions. These are called dark positions.
(2) The intersection of non-empty inclusion-monotonic sets like infinite endsegments E(k) = {k, k+1, k+2, ...} is not empty. Every non-empty endsegment shares at least one natural number with all non-empty endsegments. In fact every infinite endsegment shares infinitely many natural numbers with all infinite endsegments. Otherwise there would be a first endsegment sharing less natural numbers with its predecessors. This cannot happen, if all endsegments are infinite.
But according to ZFC, the intersection of all endsegments is empty. Since all definable endsegments satisfy
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
the empty intersection cannot be accomplished by merely definable endsegments
∩{E(k) : k ∈ ℕ_def} =/= { }.
Only by the presence of undefinable endsegments
∩{E(k) : k ∈ ℕ} = { }
can be accomplished.
(3) The simplest proof of dark natural numbers is this:
Every definable natural number k is finite and belongs to a finite set
{1, 2, 3, ..., k}.
If there are ℵo, i.e., more than any finite number, then ℕ can can only be filled and completed by dark natural numbers. This is obvious from the simple fact
∀k ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
Regards, WM