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I'm simply curious about why the following equality holds: $ \displaystyle\prod_{n\lt\omega}\aleph_n=\aleph_\omega^{\aleph_0}. $

Much thanks!

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up vote 6 down vote accepted

The following is a theorem of Tarski: $$\prod_{\alpha<\kappa}\lambda_\alpha = \bigg(\sup\{\lambda_\alpha\mid\alpha<\kappa\}\bigg)^\kappa$$

From which it follows trivially that $\prod_\omega\aleph_n = \aleph_\omega^{\aleph_0}$. Simply let $\kappa=\omega$ and $\lambda_n=\aleph_n$, since $\sup\{\aleph_n\mid n<\omega\}=\aleph_\omega$ we have that the the product of the $\aleph_n$'s equals to $\aleph_\omega^{|\omega|}=\aleph_\omega^{\aleph_0}$.

The proof of the theorem can be found in details in various books (currently I am going through Introduction to Cardinal Arithmetics, it appears as Lemma 1.6.15).

The proof in a nutshell:

First let $\mu=\sup\{\lambda_\alpha\mid\alpha<\kappa\}$. The idea behind the proof is to consider a partition of $\kappa$ into $\{A_\xi\mid\xi<\kappa\}$ (simply by the fact that $\kappa\cdot\kappa = \kappa$), then consider the product over all the parts, that is: $$\mu^\kappa = \prod_{\alpha<\kappa}\mu \le\prod_{\xi<\kappa}\ \ \prod_{\alpha\in A_\xi}\lambda_\alpha=\prod_{\alpha<\kappa}\lambda_\alpha\le\prod_{\alpha<\kappa}\mu=\mu^\kappa$$

The first $\le$ sign comes from the distributivity of products over such partitions of the index set, and the fact that for every $A_\xi$ we have that $\sup A_\xi=\mu$.

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What is $\lambda_\alpha$? – goblin Apr 30 '14 at 20:49
Just some cardinal? – Asaf Karagila Apr 30 '14 at 20:50
Wait let me think about it some more. – goblin Apr 30 '14 at 20:58
Take your time. I'm going to sleep. – Asaf Karagila Apr 30 '14 at 20:59
Does the analogue of this amazing result hold for infinite sums? i.e. $$\sum_{\alpha<\kappa} \lambda_\alpha = \kappa \cdot \left(\sup_{\alpha<\kappa}\;\lambda_\alpha\right)$$ – goblin May 1 '14 at 0:28

It is obvious that $\prod_{n<\omega}\aleph_n \preceq (\aleph_\omega)^\omega$, since the LHS is a subset of the RHS.

For the other direction, let's try to construct a injection from $(\aleph_\omega)^\omega$ (that is, the set of infinite sequences of elements of $\aleph_\omega$) into $\prod_{n<\omega}\aleph_n$. Without loss of generality we can assume that $0$ does not occur in the original sequence (say, by adding $1$ to every finite element). Now $\aleph_\omega = \bigcup_{n<\omega} \aleph_n$, so for each $\alpha\in\aleph_\omega$ there is a least $N<\omega$ such that $\alpha\in\aleph_n$ for all $n\geq N$. Thus, for each $0$-free infinite sequence of elements of $\aleph_\omega$, we can pad that sequence with $0$'s such that each element comes late enough to be in $\aleph_n$ for $n$ being its index. The nonzero elements of the padded sequence are exactly the elements of the original sequence, and in the same order. So we have an injection from $(\aleph_\omega)^\omega$ to $\prod_{n<\omega}\aleph_\omega$.

Now apply Cantor-Bernstein.

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Thanks Henning. – MediumPace Dec 12 '11 at 10:42

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