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I think that, without countable choice, I can prove quite easily that for a sequence $(S_n)_{n \in \mathbb{N}}$ of sets $S_n$ with $|S_n|=2$, the Cartesian product $\,\prod_{n=1}^\infty S_n\,$ is either empty or else has the same cardinality as $\{0,1\}^\mathbb{N}$. Namely, if $\,\prod_{n=1}^\infty S_n\,$ is non-empty then we can take an element $(\tilde{s}_n)_{n \in \mathbb{N}} \in \prod_{n=1}^\infty S_n$ and then define the bijection

\begin{align*} \prod_{n=1}^\infty S_n &\to \{0,1\}^\mathbb{N} \\ (s_n)_{n \in \mathbb{N}} &\mapsto (\delta_{s_n \, , \, \tilde{s}_n})_{n \in \mathbb{N}}. \end{align*}

But now my question is: Is it a theorem of ZF (without countable choice) that for any sequence $(S_n)_{n \in \mathbb{N}}$ of sets $S_n$ with $2 \leq |S_n| < \infty$, the Cartesian product $\,\prod_{n=1}^\infty S_n\,$ is either empty or else has the same cardinality as $\{0,1\}^\mathbb{N}$?

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    $\begingroup$ I doubt it. Suppose $(X_n)_{n\in\mathbb N}$ is a sequence of nonempty finite sets with empty product. Let $Y_n=X_1\times\cdots\times X_n$; then $(Y_n)_{n\in\mathbb N}$ is a sequence of nonempty finite sets such that no infinite subsequence has nonempty product. Let $S_n=Y_n\cup\{0\}$; then $2\le|S_n|\lt\infty$ and $\prod_{n\in\mathbb N}S_n$ is nonempty but can't have the cardinality of the continuum, being a countable union of finite sets. $\endgroup$
    – bof
    Commented Jun 5 at 23:53
  • $\begingroup$ @bof Really nice argument! Is it a theorem of ZF that a countable union of finite sets cannot have the cardinality of the continuum? (With countable choice it's easy to see that a countable union of finite sets is countable; but without countable choice, I don't know what happens.) $\endgroup$
    – Julian Newman
    Commented Jun 6 at 1:33
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    $\begingroup$ @bof Can you please explain why $\pi_{n\in\N} $ is a countable union of finite sets. $\endgroup$
    – Joel Adler
    Commented Jun 6 at 20:39
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    $\begingroup$ @JoelAdler Let $$P=\prod_{n\in\mathbb N}S_n$$. Then $$P=\bigcup_{m\in\mathbb N}\{x\in P: x_n=0\text{ for all }n\gt m\}$$ and $$|\{x\in P: x_n=0\text{ for all }n\gt m\}|=|S_1\times\cdots\times S_m|\lt\infty.$$ $\endgroup$
    – bof
    Commented Jun 6 at 23:44
  • $\begingroup$ @bof: Clever, thank you! $\endgroup$
    – Joel Adler
    Commented Jun 7 at 9:09

1 Answer 1

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No. Take for instance a sequence of two-element sets $(A_i)_{i\in\omega}$ such that every infinite subsequence has empty product ("Russellian socks"), and let $B_i=A_i\sqcup\{\{i\}\}$. Then the product $\mathcal{B}$ of the $B_i$s is nonempty, but every element of $\prod\mathcal{B}$ is eventually the identity function. It's easy to check that this means $\vert\prod \mathcal{B}\vert\not=\mathfrak{c}$.

On the other hand, a nonempty product of an infinite sequence of two-element sets must have size continuum (fixing an element of the product gives a completely explicit bijection with $2^\omega$). Moreover, if $\mathcal{B}=(B_i)_{i\in\omega}$ is a sequence of three-element sets whose product is nonempty, then either $\prod\mathcal{B}$ bijects to the continuum or there is a sequence of two-element sets $\mathcal{A}=(A_i)_{i\in\omega}$ with $A_i\subset B_i$ and $\prod \mathcal{A}=\emptyset$ (fix $f\in\prod \mathcal{B}$ and let $A_i=B_i\setminus\{f(i)\}$).

Going out on a limb, I suspect that - at least if the finite sets have size bounded strictly below $\omega$ - either the product has size continuum or we can "embed a Russellian phenomenon" somehow. But I don't know how to state this precisely right now, let alone prove it.

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    $\begingroup$ Thanks. To clarify about the first bit: don't you need $(A_i)$ to have the property that the product of every infinite subsequence is empty [rather than just that the product of the entire sequence is empty], as exemplified in the construction in bof's comment under my question? $\endgroup$
    – Julian Newman
    Commented Jun 6 at 1:41
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    $\begingroup$ @JulianNewman Yes, that's quite right - I was sloppy. Fixed! $\endgroup$
    – Noah Schweber
    Commented Jun 6 at 2:15

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