# What is the product of all nonzero, finite cardinals?

To be specific, why does the following equality hold? $$\prod_{0\lt n\lt\omega}n=2^{\aleph_0}$$

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As a product of cardinals, yes:

$$2^{\aleph_0} \leq \prod_{0 < n < \omega} n \leq {\aleph_0}^{\aleph_0} \leq 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0}$$

As a product of ordinals, no:

$$\prod_{0 < n < \omega} n \leq \prod_{0 < n < \omega} \omega = {\omega}^{\omega}$$ but the ordinal ${\omega}^{\omega}$ is countable.

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The ordinal product of finite numbers is simply $\omega$. –  Asaf Karagila Dec 10 '11 at 15:23
Yes.${}{}{}{}{}$ –  sdcvvc Dec 10 '11 at 15:24
How do you know off the bat that $2^{\aleph_0} \leq \prod_{0 < n < \omega} n$? –  pookie Dec 10 '11 at 15:26
@pookie: This is exactly the argument given in my answer. You take a product of smaller cardinals over the same index set. –  Asaf Karagila Dec 10 '11 at 15:27
To clarify, does it actually hold for a product of ordinals? The answer and comments look like they conflict a little. –  pookie Dec 10 '11 at 15:38
If $\displaystyle f\in\prod_{n\in\omega} 2$ then $f(n)\in\{0,1\}$, and in particular for $n>1$ we have that $f(n)\in n$. Therefore this is a proper subset of $\displaystyle f\in\prod_{0<n<\omega} n$, therefore the cardinality is at least continuum.
On the other hand $\omega^\omega$ has cardinality continuum, and the same argument shows that the product is a subset of $\displaystyle\prod_{n\in\omega}\omega$