# any denumerable set has uncountably many subsets

• Jun 27th 2011, 06:18 AM
hmmmm
any denumerable set has uncountably many subsets
How do I go about proving:

any denumerable set has uncountably many subsets.

I have a proposition saying that for any set A $\displaystyle |A|<|\mathcal{P}(A)$ does this help in anyway?

It is just stated as a corollary but it has not been proven.

Thanks for any help

As $\displaystyle |\mathcal{P}(A)|>|A|$ then no bijection exists between the two and so no bijection exits between $\displaystyle \mathcal{P}(A)$ and $\displaystyle \mathcal{N}$

(as there is a bijection between $\displaystyle \mathcal{N}$ and A)

so $\displaystyle \mathcal{P}(A)$ is uncountable
• Jun 27th 2011, 06:37 AM
Plato
Re: any denumerable set has uncountably many subsets
You could slightly reorder some of those statements to make is 'flow' better. However, you do have the correct ideas there.
• Jun 27th 2011, 06:49 AM
MoeBlee
Re: any denumerable set has uncountably many subsets
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.