Use my proofs at your own risk. I may be wrong. Please notify me of error.
Remarks
u ≈ v says u is equinumerous with v.
The cross product A X A is also written A2.
For the sake of completeness, if that is wanted, we note that for A finite, the matrix A X A is rarely equinumerous with A. For A ≈ n e N (the Natural Whole Numbers), A X A ≈ n2 in N such that n <= n2.
The lower case e is used for the element symbol.
0.
If A infinite, whether denumerable or not, A X A ≈ A.
To prove
1.
A X A may be represented as a matrix, such as the case of Quad I
of the Complex plane, in which a complex point may be represented by 2 partially ordered reals.
It is the sets that are important;
the matrix picture gives a "false visualization" typical of mathematics,
as in the false visualization of a line in the Euclidean plane.
2.
By the Axiom of Choice, any set can be put into a well-ordering.
That is, any subset has a least element. Thence, we have x < y or y < x or x = y.
3.
We call Ri a row of the matrix such that i is the i-th well-ordered element of A. Note that every pair in Ra
has the coordinates 〈a,y〉 such that a is held constant
and y permitted to vary 1-to-1 with x e A.
In fact, y gives the column coordinate such that y is the y-th member of the well-ordering of A.
4.
Ra ≈ A.
Follows from 〈a,y〉as described in 3.
5.
Ra ≈ Rb
Follows from 〈b,y〉with b constant, making a straightforward 1-to-1 correspondence.
6.
∪Ri = A2
Self-evident.
The subscript i is a well-ordered
member of A, and so denumerability is not implied.
7.
Any finite subset of ∪Ri ≈ A.
Follows from 5.
8.
∪Ri = ∪Xn, where X is a finite subset of ∪Ri and n e N
Exists by definition.
9.
(All n,m e N) (Xn ≈ Ra ≈ Xm).
Follows from 5. and 8.
10.
∪Ri ≈ Ra
Follows from 9.
11. Ra ≈ A
Substitution.
12.
A ≈ A2
Substitution.
13.
An ≈ A
Follows from induction on A and n.
(14.
Aℕ\0 ≈ A
We can use transfinite induction, not defined here, for 13.)
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