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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?

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Hint: $R$ and $P(N)$ have the same cardinal

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

and $Card(R)<Card(P(R))$ since for every set $E$ $Card(E)<Card(P(E))$.

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