# Calculate the cardinal of $\prod_{0<\alpha<\omega_1}\alpha$

My guess is that the cardinal $$\prod_{0<\alpha<\omega_1}\alpha$$ can be computed in the following way:

$$\prod_{0<\alpha<\omega_1}\alpha=\aleph_0\prod_{\omega\le\alpha<\omega_1}\alpha=\aleph_0\,\aleph_0^{|\{\alpha|\,\omega\le\alpha<\omega_1\}|}=\aleph_0\,\aleph_0^{\aleph_1}=\aleph_0\,2^{\aleph_1}=2^{\aleph_1}$$

Because the set $$\{\alpha|\,\omega\le\alpha<\omega_1\}=\omega_1\setminus\omega$$, and it has cardinal equal to $$\aleph_1$$ (since it is the cardinal of $$\omega_1$$, and $$|\omega|=\aleph_0<\aleph_1=|\omega_1|$$)

Is my calculation correct? did I miss something? Any suggestion about the procedure?

Thanks in advance for your time.

PS: my calculation is based on the following proposition: let $$\kappa$$ be a cardinal and let $$(\kappa_i)_{i\in I}$$ be a family of cardinals such that $$\kappa_i=\kappa$$, for all $$i\in I$$. Then:

$$\prod_{i\in I}\kappa_i=\prod_{i\in I}\kappa=\kappa^{|I|}$$

## 1 Answer

Your first equality is not correct. Well, it happens to be correct, but what I think you're doing there is rewriting your product as $$(\prod_{0<\alpha<\omega}\alpha)\cdot(\prod_{\omega\le\alpha<\omega_1}\alpha)$$ and then replacing the first product with $$\aleph_0$$. However, the first product is $$2^{\aleph_0}$$, not $$\aleph_0$$.

• The issue here is a mix-up between cardinal and ordinal multiplication and notation. The question is about the cardinality of the Cartesian product of a collection of ordinals viewed as sets, which - annoyingly - corresponds to cardinal, as opposed to ordinal, multiplication. We could more clearly, if less succinctly, ask about the cardinality of the set of functions from $$\omega_1\setminus\{0\}$$ to $$\omega_1$$ satisfying $$f(\eta)<\eta$$ for all $$0<\eta<\omega_1$$.

This winds up not making a difference in the final answer, of course - $$2^{\aleph_0}\cdot (\aleph_0)^{\aleph_1}$$ is the same as $$\aleph_0\cdot (\aleph_0)^{\aleph_1}$$ - but it is still an important issue to have clearly. In particular, it makes the final argument not quite trivial: instead of just folding one factor of $$\aleph_0$$ into $$\aleph_1$$ factors of $$\aleph_0$$, we have to write $$2^{\aleph_0}\cdot(\aleph_0)^{\aleph_1}=2^{\aleph_0}\cdot 2^{\aleph_1}=2^{\aleph_0+\aleph_1}=2^{\aleph_1}.$$

Besides this issue, however, your answer is correct: the "after $$\omega$$" component of the full product dominates the rest, and it is easier to analyze all at once.

Alternatively, we could have argued as follows: we have $$\prod_{0<\alpha<\omega_1}\alpha\le\prod_{0<\alpha<\omega_1}\omega,$$ since each $$\alpha<\omega_1$$ has cardinality $$\le\omega$$ and the cardinality of the product depends only on the cardinality of the factors. But this second product is exactly $$(\aleph_0)^{\aleph_1}=2^{\aleph_1}$$. And $$\prod_{0<\alpha<\omega_1}\alpha\ge(\aleph_1)^{\aleph_0}$$ is easy to show, so we're done.

Incidentally, it's worth noting that there's a subtle axiom of choice issue here. When we calculate this product, we're implicitly fixing - for each $$\omega\le\alpha<\omega_1$$ - a bijection $$b_\alpha: \alpha\cong\omega$$. But the existence of a family of such bijections is not provable in ZF alone. In general, "large" products don't behave well in ZF alone - indeed, the axiom of choice is really just the statement that every product of nonempty sets is nonempty (= has cardinality $$>0$$)!

• And yes, fine, the "cardinality" of a set is a bit trickier to define in the absence of choice - see here for some details - but meh.
• Yeah, the mistake in the first step was pretty lame. It's obvious when you rewrite the product as $1\times 2\times 3\times\dots$ and you think about that set as what it really is. Apr 27 '19 at 19:23
• @Akerbeltz How do you justify your first "$\le$"? The obvious comparison gives a "$\ge$" instead. Apr 27 '19 at 19:36
• Now it is just right: $\prod_{0<\alpha<\omega_1}\alpha\le\prod_{0<\alpha<\omega_1}\omega_1=\aleph_1^{\aleph_1}=2^{\aleph_1}$ Apr 27 '19 at 19:39
• @Akerbeltz Yup, that works. Apr 27 '19 at 19:43