# What are the factors of $\aleph_0$?

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?

• 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 $0$. – vadim123 Apr 28 '13 at 16:12
• 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. – Andrés E. Caicedo Apr 28 '13 at 16:17
• 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.) – Michael Hardy Apr 28 '13 at 17:03
• @Michael, $\aleph$ is not exotic at all. It was one of the first letters I have learned to write! :-) – Asaf Karagila Apr 28 '13 at 17:03
• "Exotic" is a relative term. – Michael Hardy Apr 28 '13 at 17:04

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 $\sf ZFC$.