What are some sets that, by construction, are known to have cardinalities larger than $\mathbb{R}$?

An example of what I'm looking for: the set of topologies on $\mathbb{R}$ has a cardinality of $2^{2^{2^{\aleph_{0}}}}$.

I don't want to involve CH or anything, just sets with beth number cardinalities such as that example.

  • $\begingroup$ Well, if $\mathfrak{a}$ is a cardinality such that $\mathfrak{c}\le\mathfrak{a},$ then we know that $\mathfrak{c}<2^\mathfrak{a},$ by transitivity and Cantor's Diagonalization argument. I'm not sure that this answers your question, though. $\endgroup$ Jul 25 '18 at 0:27
  • $\begingroup$ It does not. I'm looking for sets that can be described in a normal way (EG, the set of topologies on $\mathbb{R}$), not sets that are literally just described as the power set of some other set. $\endgroup$ Jul 25 '18 at 0:34
  • 1
    $\begingroup$ Well the set of subsets of $\mathbb R$ is a natural thing, is it not? It's more interesting (in my opinion) that some things that are naively the same 'type of thing' have different cardinalities. Like the set of Lebesgue measureable subsets of $\mathbb R$ has cardinality $2^c$ whereas the set of Borel measurable sets only have cardinality $c.$ A third order thing like the set of all topologies would be expected to have a cardinality of $2^{2^c},$ so I don't see what's notable about that... I'd be more excited if it were smaller. $\endgroup$ Jul 25 '18 at 0:42
  • $\begingroup$ (Or maybe the question is really about mathematical concepts that are most naturally phrased in third or higher order logic.) $\endgroup$ Jul 25 '18 at 0:46
  • $\begingroup$ Lebesgue measurable sets is a fine example. Just any type of example of any object that has some sort of descriptiom/interest in its own right. Basically, anything other than a set constructed simply with the intent of having a large cardinality (ie, P(P(P(N)))). $\endgroup$ Jul 25 '18 at 0:54

The set of functions $\mathbb{R} \to \mathbb{R}$ is $2^{2^{\aleph_0}}$. If you consider operations like integration and differentiation to be functions on functions then the set of all these second order functions to be $2^{2^{2^{\aleph_0}}}$. You can of cause run this process of taking powersets without bound.


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.