Countably infinite Cartesian product of finite sets is infinite Say that $A \neq \emptyset$, but $\emptyset \in A$. With the term "infinite set" I mean uncountable or countably infinite. I want to prove that
$A \times A \times \cdot \cdot \cdot A \times \cdot \cdot \cdot$
is an infinite set either if $A$ is finite or infinite. If A is countably infinite, I now how to proceed and I now that is true that the infinite Cartesian product is infinite. If A is finite, I simply would  show that the cardinality of the above set is $|A| * |A| * \cdot \cdot \cdot$, so basically infinite. Am I right or am I missing something? Three questions (provided that my intuition is true):

*

*if this result is true, does it hold even if $A$ has only two elements (as inferred by the assumptions)?

*If $A$ is finite, the countably infinite Cartesian product is countably infinite or uncountable? I think it is countably infinite.

*What about the case of $A$ uncountable? I think the product should be also uncountable.

Thanks.
 A: Your set:
$$
A\times A \times \ldots \times A \times \ldots
$$
can be written as $A^{\mathbb{N}}$ (the set of functions from $\mathbb{N}$ to $A$).
The cardinality of $A^{\mathbb{N}}$ is $|A|^{|\mathbb{N}|}$. This cardinality is infinite if and only if $|A|\geq 2$ (and in this case it is uncountably infinite).
So:

*

*If $A=\emptyset$ then $A^{\mathbb{N}}=\emptyset$.


*If $|A|=1$ then $|A^{\mathbb{N}}|=1$.


*If $|A|\geq 2$ then $A^{\mathbb{N}}$ is uncountable.
The proof of (3) is a standard/famous exercise using Cantor's diagonal argument. Note that it is enough to prove (3) in the case that $|A|=2$. This is because if $|A|\geq 2$ then, letting $A_{0}$ be any $2$-element subset of $A$, observe that $A_0^{\mathbb{N}} \subseteq A^{\mathbb{N}}$.
For the case $|A|=2$ you may as well assume $A=\{0,1\}$. This puts $A^{\mathbb{N}}$ in 1-1 correspondence with the powerset of $\mathbb{N}$ (identify a subset of $\mathbb{N}$ with its characteristic function). The proof that the powerset of $\mathbb{N}$ is uncountable (using Cantor's diagonal argument) is worth trying on your own, if you haven't already. It's also easy to find in this network.
