Is the powerset of the reals any "more uncountable" (in some sense) than the reals are?

Question

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 1

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.

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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.

Answer 2

Yes. Look at Cantor's theorem. Basically the proof is very similar to the real case, since it uses a diagonalization argument, although it may appear in a slightly different fashion than you are use to. For a set $A$ consider a function $f:X \to 2^X$, and the set $\{x \in X \mid x \not \in f(x) \}$. You can show that this element can not be in the image, hence $f$ is not a bijection.

Answer 3

Here is a different aspect that wasn't covered by previous answers, in the context of $\sf ZF$, rather than $\sf ZFC$.

The natural numbers are well-ordered in their natural order. This means that being countable means that you are necessarily well-orderable.

We can show that the power set of a well-orderable set can be linearly ordered, without using the axiom of choice. So $\Bbb R$ is linearly ordered without needing to appeal to any choice related principles.

But we can show that it is consistent that $\mathcal P(\Bbb R)$ cannot be linearly ordered at all.

In this aspect $\mathcal P(\Bbb R)$ is "more uncountable" than $\Bbb R$.