2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Now that makes the question clear $\endgroup$ – user210387 May 28 '15 at 4:31
  • $\begingroup$ If I'm getting the gist of the question right, one idea might be: we can form $\mathbb{R}$, an uncountable set, from the closure of $\mathbb{Q}$, a countable set; but can we form $\mathcal{P}(\mathbb{R})$ in such a way (by some extension of a countable set) as well? $\endgroup$ – Brent May 28 '15 at 4:57
  • $\begingroup$ I'm afraid I don't know what closures are yet. Are you perchance referring to how the reals can be "constructed" from the rationals? If so, yes, that would be a very interesting "property" to lose: constructibility from a countable set. $\endgroup$ – Soham Chowdhury May 28 '15 at 4:59
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ This seems like something that answers my question, but it assumes a little more knowledge about cardinals than what I possess at the moment. Could you please expand on the parts of your answer that relate to "large cardinals", please? $\endgroup$ – Soham Chowdhury May 28 '15 at 4:33
  • $\begingroup$ @SohamChowdhury Perhaps you saw that MartinSleziak added a link for large cardinals (and another). $\endgroup$ – PJTraill May 28 '15 at 6:58
5
$\begingroup$

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 $\Bbb R$.

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
9
  • 1
    $\begingroup$ Since the OP uses the phrase "cardinalities greater than C," I suspect he's not just asking for "larger" sets, although it is unclear then what exactly the question means . . . $\endgroup$ – Noah Schweber May 28 '15 at 4:25
  • $\begingroup$ You are right. I'll update the question with some details. $\endgroup$ – Soham Chowdhury May 28 '15 at 4:26
  • $\begingroup$ @NoahSchweber Well I think the idea is that you can not "count" the power set of a set with the set itself. The question is unclear though. $\endgroup$ – user29123 May 28 '15 at 4:29
  • $\begingroup$ @Paul, thanks for your answer, but this does not answer my question. I apologize for not having stated it clearly enough earlier. $\endgroup$ – Soham Chowdhury May 28 '15 at 4:30
  • 1
    $\begingroup$ @SohamChowdhury I just mean can be counted, or injected into the reals, similar to countable means being able to inject into the naturals. Studying infinite combinatorics is probably the place to look, as Noah mentions. I know very little, but I suspect it would be difficult to say anything interesting about $2^\mathbb{R}$ in particular (GCH being independent and all). $\endgroup$ – user29123 May 28 '15 at 4:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.