I know that $\mathbb{N}$ is countable and has cardinality $\aleph_0$, and that $\mathbb{R}$ has cardinality $2^{\aleph_0} = \text{C}$ and is uncountable.
Are sets with cardinalities greater than $\text{C}$ (like $2^{\mathbb{R}}$, for instance) "more uncountable" in some sense than the reals are?
Edit: I am familiar with the proof of the fact that there is no bijection from a set to its powerset. What I'm looking for is this: do we lose some more properties when we go from $\mathbb{R}$ to $2^{\mathbb{R}}$, like we lose countability when we go from $\mathbb{N}$ to $\mathbb{R}$? Are there any notions of "higher countability", or some sort of analog of countability, that $\mathbb{R}$ has, but which we miss when we consider the powerset of the reals?
Answer
You use the phrase "cardinalities greater than $C$," so I assume you know that Cantor's diagonal argument shows that for any set $X$, $\mathcal{P}(X)$ is strictly larger than $X$ (in that $X$ injects into $\mathcal{P}(X)$ but does not surject onto $\mathcal{P}(X)$).
Based on this, I think the real question is: what do you mean by "more uncountable"?
One possible answer is the following: there may be sets with combinatorial properties which are characteristic of extremely large objects, which "reasonable" infinite sets like the naturals and the reals cannot have. For example, measurability: a cardinal $\kappa$ is measurable iff there is a countably complete ultrafilter on $\kappa$ which is not principal. By combining "countably complete" with "nonprincipal," this is clearly an "uncountability property" if anything is!
I would guess that you would find large cardinals very interesting; and, I suspect that in general large cardinal properties provide positive answers to your question.
For a related question - given a set $X$ with cardinality in between $\omega$ and $C$, is $X$ "closer" to $\omega$ or $C$? - you should check out cardinal characteristics of the continuum.
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