# $\mathcal P \left({\mathbb{R}}\right)$ has strictly greater cardinality than $\mathcal P \left({\mathbb{N}}\right)$

Show that $\mathcal P \left({\mathbb{R}}\right)$ has strictly greater cardinality than $\mathcal P \left({\mathbb{N}}\right)$ without using the arithmetic of infinite cardinals.

This is a problem from Thomas Sibley's Foundations of Mathematics.

It is clear that $\mathcal P \left({\mathbb{R}}\right)$ has greater cardinality using the identity function. But I'm stuck at showing that they can't be the same.

I tried to show that any function from $\mathcal P \left({\mathbb{N}}\right)$ to $\mathcal P \left({\mathbb{R}}\right)$ can not possibly be onto. I got stuck here.

Then I assumed that there was a bijection between both sets, and tried to get a contradiction, but I don't really know how.

Any hints?

Hint: $$\mathbb{R}$$ and $$\mathcal{P}(\mathbb{N})$$ have the same cardinal
and $$\mathrm{card}(\mathbb{R})<\mathrm{card}(\mathcal{P}(\mathbb{R}))$$ since for every set $$E$$, $$\, \mathrm{card}(E)<\mathrm{card}(\mathcal{P}(E))$$.