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


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:

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


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.
  • $\begingroup$ 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. $\endgroup$
    – Akerbeltz
    Apr 27 '19 at 19:23
  • $\begingroup$ @Akerbeltz How do you justify your first "$\le$"? The obvious comparison gives a "$\ge$" instead. $\endgroup$ Apr 27 '19 at 19:36
  • $\begingroup$ Now it is just right: $\prod_{0<\alpha<\omega_1}\alpha\le\prod_{0<\alpha<\omega_1}\omega_1=\aleph_1^{\aleph_1}=2^{\aleph_1}$ $\endgroup$
    – Akerbeltz
    Apr 27 '19 at 19:39
  • 1
    $\begingroup$ @Akerbeltz Yup, that works. $\endgroup$ Apr 27 '19 at 19:43

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