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Is it always possible to calculate the cardinality of the set of choices functions of a collection?

In others words, assuming AC (axiom of choice), we have that an arbitrary cartesian product is non-void, given that each of its sets is non-void. Is it possible to determinate its cardinality? For example, given a collection of cardinals $\{\alpha_l\}_{l\in\lambda}$, how can we compute the cardinality of $$\prod_{l\in\lambda}\alpha_l$$ knowing the cardinality of each $\alpha_l$ and $\lambda$?

An especific case of interest would be: how many choices functions are there in $P^2(\omega)\setminus\emptyset$? Does it depends of CH (continuum hypothesis) or GCH (generalized continuum hypotesis)?

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  • $\begingroup$ Depends on what you mean by “calculate the cardinality”. For example, if each $\alpha_l$ is $2$, and $\lambda=\aleph_0$, then the product has cardinality $2^{\aleph_0}$, but we don’t know which cardinal that is in ZFC. $\endgroup$ Oct 10 '19 at 0:36
  • $\begingroup$ The cardinalityh of $P^2(\omega)$ (the power set of the power set of $\omega$) is $2^{\mathfrak{c}}$, with $\mathfrak{c}=2^{\aleph_0}$. With GCH, this would be $\aleph_2$, so I think you would get $\aleph_3$ for the cardinality of the set of choice functions. But without GCH, then who knows? $\endgroup$ Oct 10 '19 at 0:40
  • $\begingroup$ In a general case is it possible to describe the product as being in bijection with some expression involving the sets and its powersets? $\endgroup$ Oct 10 '19 at 0:42
  • $\begingroup$ Why would the set of choice functions of $P^2(\omega)\setminus\emptyset$ have a bijection with $P^3(\omega)$? $\endgroup$ Oct 10 '19 at 0:44
  • $\begingroup$ Or otherwise, why would it be $\aleph_3$ assuming GCH? $\endgroup$ Oct 10 '19 at 0:52
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We know that $P(\omega)$ has $2^{\aleph_0}=\mathfrak{c}$ elements; and that $P^2(\omega)$ has $2^{\mathfrak{c}}$ elements.

For simplicity, say $\mathfrak{d}=2^{\mathfrak{c}}$, and let $X=P^2(\omega)\setminus\varnothing$.

Then $$\left|\prod_{A\in X}A\right| = \prod_{A\in X}|A| \leq \prod_{A\in X}\mathfrak{c} = \mathfrak{c}^{\mathfrak{d}}.$$ But we also have $$\left|\prod_{A\in X}A\right| \geq \left|\prod_{B\in X,|B|=\mathfrak{c}}B\right| = \prod_{B\in X,|B|=\mathfrak{c}}|B| = \mathfrak{c}^{\mathfrak{d}}.$$

Also, $$2^{\mathfrak{d}}\leq \mathfrak{c}^{\mathfrak{d}}\leq (\mathfrak{d}^+)^{\mathfrak{d}}\leq (2^{\mathfrak{d}})^{\mathfrak{d}} = 2^{\mathfrak{dd}} = 2^{\mathfrak{d}},$$ so we get equality throughout.

So we can certainly say that the number of choice functions on $X$ is $2^{\mathfrak{d}} = 2^{2^{\mathfrak{c}}}$.

In the absence of CH we don’t even know what $\mathfrak{c}$. Even if we do, we don’t know what $2^{\mathfrak{c}}$ is, and we don’t know what $2^{2^{\mathfrak{c}}}$ is.

But if we have GCH, then $\mathfrak{c}=\aleph_1$, $2^{\mathfrak{c}}=\aleph_2$, and $2^{2^{\mathfrak{c}}} = \aleph_3$. So if you allow GCH, the set you are asking about has cardinality $\aleph_3$.

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  • $\begingroup$ How do you know that there are $\mathfrack{d}$ sets of cardinality $\mathfrack{c}$ in $X$? $\endgroup$ Oct 10 '19 at 11:51
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    $\begingroup$ @Arararararagi-kun: This is fairly standard; the cardinality of the collection of subsets of cardinality less than $|X|$ of an infinite set $|X|$ is $|X|$, so the cardinality of the collection of subsets of $X$ of cardinality $|X|$ is $2^{|X|}$. P.S.Use \mathfrak, not \mathfrack. $\endgroup$ Oct 10 '19 at 15:34
  • $\begingroup$ Oh, thank you, I didn't think about this. $\endgroup$ Oct 10 '19 at 15:59
  • $\begingroup$ P.S.: sorry for the typo, I'm using the mobile app and it doesn't render automatically, so I didn't see it. I couldn't edit my comment and correct it, and I don't know whether it is possible... $\endgroup$ Oct 10 '19 at 16:24
  • $\begingroup$ @Arararararagi-kun: You can only edit a comment in the first five minutes after posting it. $\endgroup$ Oct 10 '19 at 20:27

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