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

Question

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?

Answer 1

Hint: $\mathbb{R}$ and $\mathcal{P}(\mathbb{N})$ have the same cardinal

https://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Cardinal_equalities

and $\mathrm{card}(\mathbb{R})<\mathrm{card}(\mathcal{P}(\mathbb{R}))$ since for every set $E$, $\, \mathrm{card}(E)<\mathrm{card}(\mathcal{P}(E))$.