Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Much thanks!

share|improve this question
add comment

2 Answers

up vote 5 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 $\sup A_\xi=\mu$.

share|improve this answer
What is $\lambda_\alpha$? –  user18921 Apr 30 at 20:49
Just some cardinal? –  Asaf Karagila Apr 30 at 20:50
Wait let me think about it some more. –  user18921 Apr 30 at 20:58
Take your time. I'm going to sleep. –  Asaf Karagila Apr 30 at 20:59
Does the analogue of this amazing result hold for infinite sums? i.e. $$\sum_{\alpha<\kappa} \lambda_\alpha = \kappa \cdot \left(\sup_{\alpha<\kappa}\;\lambda_\alpha\right)$$ –  user18921 May 1 at 0:28
show 2 more comments

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.

share|improve this answer
Thanks Henning. –  MediumPace Dec 12 '11 at 10:42
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.