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Suppose you have 2 infinite numbers, say $A$ and $B$.

$A$ is an element of the hyperreals, so that $A$ is greater than every real number.

$B$ is the size of the set of natural numbers, $\aleph_0$

Does it make sense to compare $A$ and $B$? And if so, how can you compare these kind of numbers?

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That should be every real number. I also fixed it in the question. – Patrik Jul 29 '11 at 14:24
Maybe part of the issue has to see with the difference between ordinal and cardinal numbers. The relation between A and other real numbers is one of order/ordinality, not of cardinality. – gary Jul 29 '11 at 15:07
up vote 8 down vote accepted

In a nutshell: No. Hyperreal numbers are "non-sets objects" while $\aleph_0$ is essentially a notion of size for a set. While hyperreal numbers can be represented as sets (but not only, e.g. in ZF+Atoms the atoms may be given the structure of the hyperreal numbers), they are interpreted as something else, while $\aleph_0$ is a lot more "concrete" as it will always be interpreted as a set of some sort.

There are different notions of infinite numbers. There are hyperreal numbers, ordinal numbers, cardinal numbers, one can view real numbers as infinite sequences of rationals and so infinitely more accurate than rational numbers (just as infinitesimals give us the ability to be more accurate than real numbers).

These notions grew out of some place where they were needed, and sometimes these places are somewhat orthogonal or unrelated (at least not directly).

In this case, we consider cardinals vs. hyperreal numbers. The cardinalities (under the axiom of choice, without the axiom of choice this is an even bigger mess, however with somewhat surprising results - more on that later) are well ordered. This means that between $\aleph_0$ and $\aleph_1$ there are no other cardinals. There are ordinals but they are all countable as sets.

Suppose you could somehow identify an element, call it $B$, in the hyperreals, $^*\mathbb R$, to be $\aleph_0$. What is $B+1$? In cardinalities $\aleph_0+1=\aleph_0$. In the hyperreals this is impossible.

What you could say is that all the elements of the form $B\pm n$ are still "infinite numbers".

The same problem would be met if we chose instead to identify ordinal numbers, since $1+\omega=\omega$ (where $\omega$ is the ordinal representing the natural numbers), this addition is non-commutative, as well not inversible, since subtraction is even less nice than addition when applied on ordinal numbers.

A nice thing to consider: without the axiom of choice there can be cardinal numbers which are not $\aleph$-numbers (that is cannot be well ordered). It is consistent to have $2^{\aleph_0}$ cardinals so that they are ordered (by the natural ordering of cardinalities) like the real numbers. I would expect that it is possible to have that result extended to the hyperreal numbers. (Obviously we cannot expect cardinal addition and multiplication to be anything reversible)

This means that you can interpret the real numbers as cardinals. However none of them is even comparable with $\aleph_0$ (that is, none contain a countable subset).

(You might be interested in my answer here: "Homomorphism" from set of sequences to cardinals? which seems as though it may be somewhat relevant to this discussion.)

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This is a clear answer, thank you. – Patrik Jul 29 '11 at 15:08
Nice answer. Here are a couple of other rough, intuitive ways of putting it. (1) The hyperreals inherit the first-order structure of the reals, and that structure only singles out two numbers for special axiomatic roles: 0 and 1. There is no obvious way to bring in some new unit like $\aleph_0$. (2) If you look at constructive approaches to the hyperreals, they essentially involve formalizing the notion of infinite numbers as diverging sequences. There is no obvious way to match up this notion with the set-theoretical notions of cardinals and ordinals. The OP might be interest in the surreals. – Ben Crowell Jul 29 '11 at 15:42
@Ben Crowell I always was wondering how one can reconcile diverging series with ordinals. I think I know a sketch. – Anixx May 7 '15 at 6:20

These are quite different notions, so the answer I suspect most people will give you is "NO" (and I suspect this question might get closed as well). However, I would like to point out that there are a lot of situations in mathematics where seemingly unrelated notions were later brought under the umbrella of some unifying idea. For example, the convex hull of a set and the sigma-algebra generated by a collection of sets, such as was pointed out yesterday: The $\sigma$-algebra of subsets of $X$ generated by a set $\mathcal{A}$ is the smallest sigma algebra including $\mathcal{A}$ Also, there are a lot of situations in mathematics where notational similarity or other essentially non-mathematical resemblance has led to the creation of unifying ideas. For example, fractional derivatives or the idea of complex-iterates of a function.

That said, I doubt anything nontrivial will come out of trying to find a unifying idea underlying these two notions of "infinite number". The only thing that occurs to me is that, in some kind of way that is relative to the underlying "universe" each notion is embedded in, we're looking at the smallest extension outside of certain "small numbers".

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Are we that bad? :-) I don't think I've ever seen a question as well-posed and sensible as this one closed. – joriki Jul 29 '11 at 14:50
O-K, so chalk it up to me being so new here! I guess I'm not yet a good judge of what gets closed and what doesn't. – Dave L. Renfro Jul 29 '11 at 14:53

Actually, yes. You can look at this paper:

The surreal numbers class includes the number $\omega$ which is identified with the infinite ordinal $\omega$ and infinite cardinal $\aleph_0$.

But bear in mind that arithmetic operations defined on surreal numbers differ from those usually defined for ordinals and cardinals. The operations defined on surreals are equivalent to the natural operations on ordinals.

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Consider the set $(\Bbb R\setminus\{0\})\cup\{\omega\}$ and redefine $x+\omega=x$ and $x\cdot\omega=\omega$. OH MY GOD! The real numbers are comparable with $\aleph_0$!!! The point is that Surreal numbers, while extending both the reals and the ordinals/cardinals disagree with both ordinal addition and cardinal addition. $1+\omega=\omega+1$ in the Surreals, but not in ordinal arithmetic; and $1+\omega\neq\omega$ which is not the case in cardinal addition. – Asaf Karagila May 6 '15 at 21:38
@Asaf Karagila exactly my point. The operations on surreals correspond not to the classical operations on ordinals but to the so-called "natural operations" (follow the second link). – Anixx May 6 '15 at 22:02
And those correspond to cardinal arithmetic how? – Asaf Karagila May 6 '15 at 22:03
So... just take the first $2^{\aleph_0}$ cardinals, define some field structure isomorphic to the reals, and then you have a field of real numbers which also compared ordinals and the cardinals they are identify. – Asaf Karagila May 6 '15 at 22:12
You keep sending me to Wikipedia, like I'm not doing set theory for a living. Do you even read my comments? Because you seem to miss the point I was making, over and over again. – Asaf Karagila May 7 '15 at 5:34

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