Once again you ignore Cantor's construction method of P(S) members, which actually enables to define a 1-to-1 correspondence between S and P(S) members, by using this construction |P(S)| times.
Actually by Cantor's construction method:
S ↔ P(S)
Your symbolism needs some defining, coz if "↔" is supposed to mean bijection, as you mentioned "1-to-1 correspondence," then S ↔ P(S) is wrong, coz it implies |S| = |P(S)|.
Are you going to help counting? I have this set
S = {1, 2, 3, 4, ...}
and I decided to count all the subsets in P(S) with cardinality equaling 3. But it's a long count . . .
{1 2 3}, {1 2 4}, {1 2 5}, {1 2 6}, . . .
{1 3 4}, {1 3 5}, {1 3 6}, {1 3 7}, . . .
{1 4 5}, {1 4 6}, {1 4 7}, {1 4 8}, . . .
{1 5 6}, {1 5 7}, {1 5 8}, {1 5 9}, . . .
.
.
.
I interlaced the sets the way Cantor did it with rational numbers, but never got to {2 x y}. Do you know how to biject subsets of S where the cardinality of such subsets is C = 3? The only way possible is to organize the above sets in a 3-dim space. It means that in order to arrive at |
N| = |P(
N)|, you need to organize the power set in
(aleph0)!-dim space, where "!" is the factorial. The only person in the universe who could do that is of course you. Take your time . . .