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Extend the system of positive natural numbers with $\aleph_0$. Then we have:

$$\aleph_0 = \aleph_0\cdot n,\quad \forall n \in \mathbb{N}^+$$

Does it make sense to talk about factors of $\aleph_0$? What are the factors of $\aleph_0$?

Aside: Are there systems of numbers where it makes sense to talk about factors of infinite numbers?

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    $\begingroup$ You've already found the factors, but it's not a particularly useful piece of knowledge. You could also say that $0*n=0$ for all $n$, so you've found the factors of $0$. $\endgroup$ – vadim123 Apr 28 '13 at 16:12
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    $\begingroup$ There is an interesting topic lurking in the background here: Note that $\aleph_0$ is also a factor, and that the collection of factors of $\aleph_0$ is closed under finite products. But it is not closed under countable products. The question of factors on its own is not too interesting in the sense that $\kappa\lambda=\lambda$ if $\kappa,\lambda$ are nonzero cardinals and $\lambda$ is infinite. On the other hand, looking at infinite products (countable or longer) does not lend itself to such an easy answer. $\endgroup$ – Andrés E. Caicedo Apr 28 '13 at 16:17
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    $\begingroup$ Why do you use an asterisk for multiplication in $\TeX$? The reason asterisks are used is that one is limited to the symbols on the typewriter keyboard and the letter x is needed for use as a variable in programming languages. In $\TeX$ you can write $3\cdot5$ or $3\times5$ or $3\otimes5$ or lots of other things, an even exotic things like $\aleph$, so why go back to a usage invented for occasions when one has a severely impoverished set of symbols available? (I changed it in the posted question.) $\endgroup$ – Michael Hardy Apr 28 '13 at 17:03
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    $\begingroup$ @Michael, $\aleph$ is not exotic at all. It was one of the first letters I have learned to write! :-) $\endgroup$ – Asaf Karagila Apr 28 '13 at 17:03
  • $\begingroup$ "Exotic" is a relative term. $\endgroup$ – Michael Hardy Apr 28 '13 at 17:04
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No. Cardinals are not suitable for talking about decomposition and factors.

The reason is that $\kappa\cdot\lambda=\max\{\kappa,\lambda\}$. So no cardinal can be expressed as "nontrivial" finite products of smaller cardinals. For infinite products we cannot really prove much in $\sf ZFC$.

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    $\begingroup$ On the other hand, on settings where choice fails (badly) one can develop a meaningful theory of factors. See this answer: mathoverflow.net/questions/63596/… $\endgroup$ – Andrés E. Caicedo Apr 28 '13 at 18:19
  • $\begingroup$ Of course. One can also develop a meaningful theory of the additive structure of cardinals. See Monro's paper "Decomposable cardinals", Fund. Math. 80 (1973), no. 2, 101–104. $\endgroup$ – Asaf Karagila Apr 28 '13 at 18:36
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    $\begingroup$ Well, I would say that one can certainly talk about factors of infinite cardinals. It's just that the answer (as you've given) turns out to be uninteresting. $\endgroup$ – Pete L. Clark Apr 28 '13 at 19:12

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