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?
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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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0The ordinal product of finite numbers is simply $\omega$. – 2011-12-10
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0Yes.${}{}{}{}{}$ – 2011-12-10
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1How do you know off the bat that $2^{\aleph_0} \leq \prod_{0 < n < \omega} n $? – 2011-12-10
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0@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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0To 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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0@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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0Oh, 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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0@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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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'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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0I think that I meant to say that $\prod_{n\in\omega}2\subseteq\prod_{n\in\omega}n$. – 2014-05-01