Reading "Algebraic Methods in Philosophical Logic" by Dunn and Hardegree, the following claim is made on page 29:
Every direct product of finite algebras is either finite or uncountable; no such direct product is denumerable.
Surely a family of finite algebras indexed by a denumerable index set yields a denumerable direct product?
Thanks for any help or pointers!
EDIT: Here are the definitions of algebra and direct product given in the book:
An algebra $\mathbf{A}$ is defined to be an operational structure. An operational structure is, by definition, a set $A$ together with a family $\langle O_i \rangle$ of operations on $A$.
Recall that a choice function on a family $\langle A_i \rangle$ of sets is any family $\langle a_i\rangle$ of elements such that $a_j\in A_j$ for all $j$ in the (implicit) indexing set. The Cartesian prouct of the family $\langle A_i\rangle$ of sets, denoted $\times\langle A_i\rangle$, is defined simply to be the set of all choice functions on $\langle A_i \rangle$.
The direct product of algebras has as a carrier set $\times\langle A_i\rangle$ and a way to specify the operations on this Cartesian product. The detail should not be relevant as they say the reason the direct product is not denumerable is because the Cartesian product is not denumerable.
This has nothing to do with algebras: any product of finite sets is either finite or uncountable (using the axiom of choice).
This follows, as bof commented, from the fact that $\{0,1\}^\mathbb N$ is uncountable: if you have a family of nonempty finite sets $(X_i)_{i\in I}$, then either $J=\{i\in I\mid |X_i|\geq 2\}$ is infinite, in which case the product has cardinality $= \prod_{j\in J}|X_j| \geq \prod_{j\in J}2 \geq 2^\mathbb N >|\mathbb N|$, or it is finite, in which case $\prod_{j\in J}|X_j|$ is then finite.