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

2 Answers 2

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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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    @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, 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 $\displaystyle\prod_{n\in\omega}\omega$

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    I think that I meant to say that $\prod_{n\in\omega}2\subseteq\prod_{n\in\omega}n$.2014-05-01