Final answer:
To prove the countability of the union of countable sets A_n, we can use a diagonal argument to establish a one-to-one correspondence with natural numbers; the Axiom of Choice's use depends on the proof's details.
Step-by-step explanation:
In addressing the question regarding a sequence of countable sets An, we begin by confirming that the set {A0, A1, ..., An, ...} is indeed a set. The existence of this set follows from the Axiom of Specification (also known as the Subset Axiom), which states that for any set and any condition, there exists a subset that contains exactly those elements of the set that satisfy the condition. Since each An is defined and countable, grouping them together into a single set does not contradict any of the principles of set theory.
For part (b), we are required to prove that the union ∅i≥0 Ai is countable. This can be done by constructing a function that provides a one-to-one correspondence between the natural numbers and the elements of the union. We can utilize the fact that countable sets can be listed in sequence (e.g., as a sequence of elements) and interleave these sequences (diagonal argument) to form a single sequence, thus establishing countability.
For part (c), the need for the Axiom of Choice depends on the details of the proof for part (b). In typical proofs that the countable union of countable sets is countable, the Axiom of Choice might be used implicitly, especially if one selects a representative element from each sequence without providing an explicit rule. However, if each An can be well-ordered without invoking the Axiom of Choice (e.g., if we have an explicit listing or rule to pick elements), then we can perform the count without it.