# Cardinality of an arbitrary cartesian product

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)?

• 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. Oct 10 '19 at 0:36
• 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? Oct 10 '19 at 0:40
• In a general case is it possible to describe the product as being in bijection with some expression involving the sets and its powersets? Oct 10 '19 at 0:42
• Why would the set of choice functions of $P^2(\omega)\setminus\emptyset$ have a bijection with $P^3(\omega)$? Oct 10 '19 at 0:44
• Or otherwise, why would it be $\aleph_3$ assuming GCH? Oct 10 '19 at 0:52

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 $$\aleph_3$$.

• How do you know that there are $\mathfrack{d}$ sets of cardinality $\mathfrack{c}$ in $X$? Oct 10 '19 at 11:51
• @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. Oct 10 '19 at 15:34