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

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up vote 5 down vote accepted

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

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Thanks Sir Karagila. – pookie Dec 10 '11 at 15:28
I'm reading your answer to understand the first inequality in what sdcvvc wrote, but I don't understand the bit "Therefore this is a proper subset of f" What is a proper subset of $f$, and why are we interested in proper subsets of $f$ anyway, given that $f$ is a function? (Oh, I know that every function can be viewed as a relation, and hence that a proper subset of a function is just a a restriction of the original function, but I don't see the significance here.) – goblin May 1 '14 at 20:46
I think that I meant to say that $\prod_{n\in\omega}2\subseteq\prod_{n\in\omega}n$. – Asaf Karagila May 1 '14 at 20:48

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