Answer
See explanation
Work Step by Step
**Claim.** If \(A\) is countably infinite, \(B\) is any set, and \(g : A \to B\) is onto (surjective), then \(B\) is countable (i.e., \(B\) is either finite or countably infinite).
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## Intuitive Explanation
A surjective map \(g\colon A \to B\) means every element of \(B\) is “hit” by some element of \(A\). Since \(A\) is countably infinite, we can list out its elements as \(A = \{a_1, a_2, a_3, \dots\}\). Because \(g\) is onto:
- Every \(b \in B\) appears as \(g(a_n)\) for at least one \(n\).
- By choosing the *first* (or “smallest”) index \(n\) for which \(g(a_n)=b\), we get a way to encode each \(b\) as a *unique* natural number \(n\).
Hence there is an **injection** of \(B\) into \(\mathbb{N}\). Once you can inject \(B\) into \(\mathbb{N}\), it follows that \(B\) is at most countable. And since the surjection from a countably infinite set typically produces an infinite image (unless \(B\) is finite), we conclude \(B\) is countable.
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## Formal Sketch of the Proof
1. **List \(A\)**.
Because \(A\) is countably infinite, there is a bijection \(f\colon \mathbb{N} \to A\). Equivalently, you can label the elements of \(A\) as \(\{a_1, a_2, a_3, \dots\}\).
2. **Compose to get a surjection from \(\mathbb{N}\) to \(B\)**.
Define \(h = g \circ f \colon \mathbb{N} \to B\). This is onto \(B\) because \(g\) is onto \(B\) and \(f\) is onto \(A\).
3. **Construct an injection from \(B\) into \(\mathbb{N}\)**.
For each \(b \in B\), the surjectivity of \(h\) guarantees there is at least one \(n\) with \(h(n)=b\). Among all such \(n\), pick the smallest one and call it \(\ell(b)\). Formally,
\[
\ell(b) \;=\; \min \{\,n \in \mathbb{N} : h(n)=b\,\}.
\]
This defines a function \(\ell : B \to \mathbb{N}\).
4. **\(\ell\) is injective**.
If \(\ell(b_1) = \ell(b_2)\), that means \(b_1\) and \(b_2\) were first hit by the same smallest natural number. Hence \(h(\ell(b_1)) = b_1\) and \(h(\ell(b_2)) = b_2\) must coincide. Thus \(b_1 = b_2\). So \(\ell\) is one‐to‐one.
5. **Conclusion**.
An injective map \(\ell : B \to \mathbb{N}\) shows \(B\) is at most countable. In fact, if infinitely many elements of \(B\) are hit, \(B\) is countably infinite; if finitely many are hit, \(B\) is finite. In all cases, \(B\) is **countable**.
Hence any surjective image of a countably infinite set is itself **countable**.
