Suppose first that for every . Then an element of is determined by its first coordinates, so is in bijection with
A finite Cartesian product of countable sets is countable by repeated application of the diagonal enumeration in part (b)(i). Hence is countable.
Conversely, suppose infinitely many factors contain at least two elements. Choose increasing indices and distinct elements
Fix one element in every remaining factor. Each binary sequence then defines an element of by placing in coordinate and the fixed element elsewhere. This map is injective.
The set is uncountable by Cantor's diagonal argument: from any proposed list of binary sequences, form a new sequence whose th digit differs from the th digit of the th listed sequence. It is absent from the list. Thus 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 and while leaving free guarantees infinitely many zeros and infinitely many agreements with any prescribed binary sequence . Infinite free coordinates remain even after both requirements are enforced.
The bijections of a countably infinite set form an uncountable set. Partition an enumeration into disjoint pairs and encode an infinite binary sequence by independently fixing or swapping each pair.
If is a countably infinite set and is a finite set with at least two elements, the set of surjective functions is uncountable. Fix finitely many values to cover , then encode an arbitrary infinite binary sequence using two values of on the remaining inputs.