Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recently, I was wondering how division of aleph numbers would work. First, I thought about how finite cardinality division would work. What I came up with was that the result of $A/B$ where $A$ and $B$ are both cardinalities, is the number of times that each element of $B$ had to be mapped to an element of $A$ in order to ensure that all elements of $A$ were mapped to.

Extending this to aleph numbers, specifically, I thought that $\aleph_1/\aleph_0$ would be $\aleph_1$. The reasoning behind this is that there are an infinite ($\aleph_1$) number of real numbers between any two natural numbers. As such, that mapping would need to be applied.

Can anyone validate this idea? Is this how division of aleph numbers actually works, or am I totally off base?

Thanks.

share|improve this question
1  
Yes, $\aleph_1/\aleph_0 = \aleph_1$ is the only thing that makes sense. But what about $\aleph_0/\aleph_0$ ... shouldn't you do that one first? And what about $\aleph_0/\aleph_1$ ... –  GEdgar May 18 '12 at 22:02
add comment

2 Answers 2

up vote 8 down vote accepted

Much like I wrote in Cardinal number subtraction, if $\kappa$ and $\lambda$ are two $\aleph$-numbers, it might be possible to define division, but this definition would have to be limited and awkward.

If $\kappa$ and $\lambda$ are both regular cardinals and $\kappa<\lambda$ then every partition of $\lambda$ into $\kappa$ many parts would have to have at least one part would be of size $\lambda$. In a sense this means that $\frac\lambda\kappa=\lambda$. This is indeed the case with $\aleph_1/\aleph_0$, both are regular cardinals are $\aleph_0<\aleph_1$.

If, however, $\kappa=\lambda$ this is no longer defined, since $\kappa=2\cdot\kappa=\aleph_0\cdot\kappa=\ldots=\kappa\cdot\kappa=\ldots$, so there can be many partitions of $\lambda$ into $\kappa$ many parts, and in each the parts would vary in size (singletons; pairs; countably infinite sets; etc.)

When $\lambda$ is a singular limit cardinal, e.g. $\aleph_\omega$ this breaks down completely, since singular cardinals can be partitioned into a "few" "small" parts. In the $\aleph_\omega$ case these would be parts of size $\aleph_n$ for every $n$, which make a countable partition in which all parts are smaller than $\aleph_\omega$.

The only reasonable way I can think that cardinal division can be defined would have to consider the Surreal numbers, and the embedding of the ordinals in them. However this will not be compatible with cardinal arithmetic at all (the surreal numbers form a field).

I should also remark that your reasoning for $\aleph_1/\aleph_0$ being $\aleph_1$ is invalid. First note that neither is a real number, and that it is possible that $\aleph_1$ is much smaller than the cardinality of the real numbers (so between two natural numbers there are a lot more real numbers). Secondly, note that between two rational numbers there are also infinitely many rational numbers - does that mean $\aleph_0/\aleph_0=\aleph_0$?

However your rationale is not that far off, as I remarked in the top part of the post, if you take a set of size $\aleph_1$ and partition it into $\aleph_0$ many parts you are guaranteed that at least one of the parts would have size $\aleph_1$.


Further reading:

  1. Cofinality of cardinals
  2. Cofinality and its Consequences
  3. How to understand the regular cardinal?
  4. How far do known ordinal notations span? (Cantor normal form)
  5. surreal and ordinal numbers
share|improve this answer
    
Very nice. Do we need some sort of choice principle to determine that, given a partition of $\aleph_1$ into $\aleph_0$-many parts, at least one of the parts must have size $\aleph_1$? I understood that it is consistent with ZF that $\aleph_0$ is the only regular cardinal, so doesn't that mean that, for example, $\aleph_1$ can be realized as a countable union of countable subsets? –  Cameron Buie May 18 '12 at 23:25
    
@Cameron: Of course. I assume choice through and through in this answer. Indeed it is consistent with ZF (and without large cardinals at all) that $\aleph_1$ is a countable union of countable sets. If you want two successive singular cardinals you need some pretty large cardinals in the background, though. –  Asaf Karagila May 18 '12 at 23:39
add comment

If you're looking for something compatible with cardinal multiplication, you'll have to deal with the fundamental problem that $\kappa\cdot\lambda=\max\{\kappa,\lambda\}$ whenever $\kappa,\lambda$ are well-orderable cardinals and at least one of them is an aleph. That sort of absorption means--for example--that $\aleph_0\cdot\aleph_2=\aleph_1\cdot\aleph_2=\aleph_2\cdot\aleph_2=\aleph_2$, so even trying to define $\aleph_2/\aleph_2$ in some way compatible with cardinal multiplication is problematic. Now, one could choose a convention for $\lambda/\kappa$ in instances that $\kappa\leq\lambda$--say, for example, that it's always just $\lambda$--but trying to compatibly define it when $\kappa>\lambda$ is fruitless.

share|improve this answer
    
What is the difference between being well-orderable and being aleph? I was under the impression that the aleph numbers are defined as well-orderable cardinalities. –  Dustan Levenstein May 19 '12 at 2:14
    
Apologies for the ambiguity. The definition of aleph that I learned was infinite well-orderable cardinalities (i.e: $\aleph_0$ is the least aleph). Does that clarify things sufficiently? –  Cameron Buie May 19 '12 at 3:17
    
oh, yes, of course. I wasn't even thinking about finite cardinals. –  Dustan Levenstein May 19 '12 at 12:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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