You could slightly reorder some of those statements to make is 'flow' better. However, you do have the correct ideas there.
How do I go about proving:
any denumerable set has uncountably many subsets.
I have a proposition saying that for any set A does this help in anyway?
It is just stated as a corollary but it has not been proven.
Thanks for any help
As then no bijection exists between the two and so no bijection exits between and
(as there is a bijection between and A)
so is uncountable
You don't even have to involve the cardinality operator '| |'.
We'd have already proven that there is no bijection between A and PA (by Cantor's argument that there is no surjection from A onto PA).
So if A is denumerable (countably infinite), then PA is not denumerable. And PA is infinite.
Therefore, PA is uncountable.