To be specific, why does the following equality hold? $ \prod_{0\lt n\lt\omega}n=2^{\aleph_0} $
What is the product of all nonzero, finite cardinals?
6
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set-theory
cardinals
2 Answers
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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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0@sdcvvc: I edited in the word "ordinal" to make it absolutely clear, I hope you don't mind. – 2011-12-10
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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
On the other hand $\omega^\omega$ has cardinality continuum, and the same argument shows that the product is a subset of $\displaystyle\prod_{n\in\omega}\omega$
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0I think that I meant to say that $\prod_{n\in\omega}2\subseteq\prod_{n\in\omega}n$. – 2014-05-01