# Are there number systems corresponding to higher cardinalities than the real numbers?

As most of you know, the set $\omega$ with cardinality $\aleph_0$ corresponds to what we normally know as the natural numbers $\mathbb{N}$, and the set $\mathcal{P}(\omega)$ with cardinality $\aleph_1$ corresponds to what we call the real numbers $\mathbb{R}$.

Now, my question is whether there are currently known any number systems which we actually have some rough understanding of that correspond to higher cardinalities than $\mathbb{R}$? Of course, it is easy to construct a set with cardinality $\aleph_2 > \aleph_1$ by taking the power set again $\mathcal{P}(\mathcal{P}(\omega))$, but can we get any grasp on what kind of systems might have this cardinality? And if such a number system exists, how far up the chain of cardinalities do we have to go before we longer have any understanding?

• There are some nice algebraic extensions of $\mathbb R$, like the hyper-reals, but I think the hyper-reals only have size $\mathfrak c$... Nov 23, 2014 at 19:59
• If just completeness(every Cauchy sequence is convergent) of a number system is important for someone, then every complement of $\mathbb{Q}$ is isomorphich to $\mathbb{R}$ ! Nov 23, 2014 at 20:01
• What's the cardinality of something like $\mathsf M_4(\mathbb R)$? Still $\mathfrak c$? Nov 23, 2014 at 20:03
• This question is much easier to answer assuming $\neg\sf CH$ - would such an answer be acceptable? Nov 23, 2014 at 20:04
• @TonyK: I suppose that the meaning was completion, not complement? Nov 23, 2014 at 20:09

## 2 Answers

First of all, let me clear a confusion of yours.

$\Bbb R$ has cardinality $2^{\aleph_0}$. This is not necessarily $\aleph_1$. It is consistent with the axioms of modern set theory that the answer is positive or negative. Similarly $2^{2^{\aleph_0}}$ need not be equal to $\aleph_2$, and it could be much larger.

Now that we cleared that issue. What is a number system? If you mean a field, simply something which has a structure of addition and multiplication like we know on $\Bbb Q$ or $\Bbb R$ or so on, then take any set $X$, and consider the ring of polynomials $\Bbb R[X]$ whose indeterminate variables come from $X$. Then $\Bbb R[X]$ is an integral domain, and therefore it has a fraction field, which is both a field and has the cardinality of $\max\{|\Bbb R|,|X|\}$. Picking a suitable $X$ would do.

If by a number system you mean something like ordinals, which are numbers (at least by virtue of being called "ordinal numbers") then ordinals have four natural operations, each defined from the previous one by transfinite induction: successorship, addition, multiplication and exponentiation.

Moreover we can show that if $\alpha$ and $\beta$ are infinite ordinals, then regardless to whatever operation you use on $\alpha$ and $\beta$, the cardinality of the result is the maximal one between $\alpha$ and $\beta$. This is wonderful, since it means that we don't have to look very far to find ordinals $\gamma$ with the property that $\{\alpha\mid\alpha<\gamma\}$ is closed under all operations.

For example, every infinite ordinal which is an initial ordinal (the least one of a given cardinality) is closed under all these operations. So $\omega_1$ and $\omega_2$ and $\omega_{\omega_1^\omega\cdot 5+\omega^{42}+1}$ are all such ordinals (recall that $\omega_\alpha$ denotes the $\alpha$-th initial ordinal). Pick any such ordinal whose cardinality is larger than the cardinality of the continuum, and you have an example.

It might be worth noting that the proper class of all ordinals is another such example, and it's not even a set!

On the other side of set theoretic number systems we have the cardinals, which again are called cardinal numbers, and these represent the size of a set. They also have the same four operations as ordinals, but now we can't define them from one another. We say that a cardinal $\kappa$ is a strong limit cardinal if whenever $\lambda<\kappa$ we have that $2^\lambda<\kappa$. This implies, amongst other things, that if $\kappa$ is a strong limit cardinal then we can show that the set of all cardinals below $\kappa$ (including the finite ones, of course) is closed under all operations on the cardinals, which will also make a number system.

And again, it might be worth noting that the class of all cardinals (what Cantor called Tav, the last letter of the Hebrew alphabet) is an example similar to the class of ordinals. It's a number system which is too large to even be a set.

• Would this construction hold regardless of whether one assumes the continuum hypothesis to be true?
– mrp
Nov 23, 2014 at 20:08
• Of course. Note that I didn't refer to that in the second part of my answer. Nov 23, 2014 at 20:09
• If you are looking for a "natural number system", then there's nothing really, since most of "natural" things are ultimately related to the real numbers. Of course if by number systems you mean other things, like ordinals or cardinals or things that behave more like the natural numbers, then we can have a whole other discussion. Otherwise, yes, this is really just cheating because that's essentially the only way (we can cheat in other ways, but those amount to artificial inflations of some field). Nov 23, 2014 at 20:21
• Ordinals and cardinals are definitely allowed.
– mrp
Nov 23, 2014 at 20:25
• @tomasz: Only if you buy into all those forcing axioms. Why not a real-valued measurable? Way more fun, and way larger. Nov 25, 2014 at 4:40

The surreal numbers are an ordered field too big even to be a set: they form a proper class that includes the ordinals.

• A fine answer, Sir! :) Nov 23, 2014 at 20:15
• @gnometorule: It’s a good thing that I had just set down my coffee cup! :-) Nov 23, 2014 at 20:17
• @Brian M. Scott, What??? Jul 28, 2021 at 13:31
• @RounakSarkar: I’m not sure what you’re asking. Jul 28, 2021 at 18:12
• @Brian M. Scott. Why are you talking about setting down your coffee cup? (You didn't expect someone to reply to a comment you posted 6 years ago right) Jul 30, 2021 at 5:51