Answer
See explanation
Work Step by Step
A standard proof uses the fact that each countably infinite set can be **listed** (put in one‐to‐one correspondence with the natural numbers). Let \(A\) and \(B\) be two countably infinite sets. Then:
1. **Enumerate \(A\)**: Since \(A\) is countably infinite, there is a bijection
\[
f\colon \mathbb{N}\;\to\;A.
\]
In other words, we can list all elements of \(A\) as \(a_1, a_2, a_3, \dots\)
2. **Enumerate \(B\)**: Similarly, there is a bijection
\[
g\colon \mathbb{N}\;\to\;B.
\]
So we can list all elements of \(B\) as \(b_1, b_2, b_3, \dots\)
3. **Interleave the two lists**: We can create a single list that includes **all** elements of \(A \cup B\) by “weaving” the two enumerations together:
\[
a_1, \; b_1, \; a_2, \; b_2, \; a_3, \; b_3, \; \dots
\]
Formally, define \(h\colon \mathbb{N}\to A\cup B\) by
\[
h(2n) = a_n
\quad\text{and}\quad
h(2n+1) = b_n.
\]
Every element of \(A\cup B\) appears somewhere in this interleaving (because each element of \(A\) appears in the even positions, and each element of \(B\) appears in the odd positions).
4. **At most countable**: Since \(\mathbb{N}\) is countable, any image of \(\mathbb{N}\) under a function (like \(h\)) can be at most countable. Thus \(A\cup B\) is at most countable.
5. **Infinite implies countably infinite**: Because both \(A\) and \(B\) are infinite, their union \(A\cup B\) is certainly infinite. An infinite set that is at most countable must be **countably infinite**.
Hence the union \(A \cup B\) of two countably infinite sets is itself **countably infinite**.
