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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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    The ordinal product of finite numbers is simply $\omega$.2011-12-10
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    Yes.${}{}{}{}{}$2011-12-10
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    How do you know off the bat that $2^{\aleph_0} \leq \prod_{0 < n < \omega} n $?2011-12-10
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    @pookie: This is exactly the argument given in my answer. You take a product of *smaller* cardinals over the same index set.2011-12-10
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    To clarify, does it actually hold for a product of ordinals? The answer and comments look like they conflict a little.2011-12-10
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    @pookie: It does not hold for ordinal product, since the product is countable (it is equal to $\omega$), while $2^{\aleph_0}$ is not countable. Mind that ordinal product and cardinal product are different operations.2011-12-10
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    Oh, I think I was thrown off since the answer seems to say that the ordinal product is $\omega^\omega$, not $\omega$, although I guess the conclusion would be the same.2011-12-10
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    @pookie: I was lazy and wrote it is at most $\omega^{\omega}$ (which also countable, as this is ordinal exponentiation), and this is enough to give the conclusion.2011-12-10
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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'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*.)2014-05-01
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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