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A set is a countable set if it admits an injective function into the natural numbers; this includes finite sets and countably infinite sets. A countably infinite set can be listed by a bijection from $\mathbb{N}$.

Let $Z$ be the set of infinite binary sequences. Suppose it were countable and list all its members as $s^{(1)},s^{(2)},\dots$, writing $s_n^{(j)}$ for the $n$th entry of row $j$. A finite listing could be repeated to make such a list as well. Define the binary sequence $y$ by $y_n=1-s_n^{(n)}$. It differs from the $n$th row in coordinate $n$, for every $n$. Thus it belongs to $Z$ but is missing from the list, a contradiction. The set $Z$ is uncountable, by Cantor's diagonal argument.

Suppose all functions $\mathbb{N}\to \mathbb{N}$ could be listed as $f_1,f_2,\dots$. DefineThen $g$ differs from $f_n$ at $n$ for every $n$, so it is absent from the list. This Cantor's diagonal argument proves that ${\mathbb{N}}^{\mathbb{N}}$ is an uncountable set.

Suppose first that $|A_n|=1$ for every $n>N$. Then an element of $B$ is determined by its first $N$ coordinates, so $B$ is in bijection withA finite Cartesian product of countable sets is countable by repeated application of the diagonal enumeration in part (b)(i). Hence $B$ is countable.

Conversely, suppose infinitely many factors contain at least two elements. Choose increasing indices $n_1<n_2<\cdots$ and distinct elementsFix one element in every remaining factor. Each binary sequence $\epsilon =({\epsilon }_1,{\epsilon }_2,\dots )$ then defines an element of $B$ by placing $a_j^{{\epsilon }_j}$ in coordinate $n_j$ and the fixed element elsewhere. This map is injective.

The set $\{0,1{\}}^{\mathbb{N}}$ is uncountable by Cantor's diagonal argument: from any proposed list of binary sequences, form a new sequence whose $j$th digit differs from the $j$th digit of the $j$th listed sequence. It is absent from the list. Thus $B$ contains an uncountable subset and cannot be countable.

Therefore

Partition the positive natural numbers into three infinite coordinate sets. Fix the coordinates on the first two sets to any prescribed binary values, and leave those on the third set free. Reading the third coordinate set embeds all infinite binary sequences in the resulting family, so the family is an uncountable set by Cantor's diagonal argument. For example, fixing $x_{3k-2}=0$ and $x_{3k-1}=a_{3k-1}$ while leaving $x_{3k}$ free guarantees infinitely many zeros and infinitely many agreements with any prescribed binary sequence $(a_n)$. Infinite free coordinates remain even after both requirements are enforced.