$A$ is a finite $\implies$ $f(A)$ is finite

Question

Please check my proof! I'm very happy to receive any suggestion to make my proof better!

Theorem:

Suppose that $A$ is a finite and that $f$ is a mapping from $A$ to $B$. Then $f(A)$ is finite.

Proof:

Since $A$ is finite, there exists a bijection $g:I_n \to A$.

$f:A \to f(A)$ is surjective $\wedge g:I_n \to A$ is bijective $\implies f \circ g:I_n \to f(A)$ is surjective.

Let $h:f(A) \to I_n$ such that $h(x)=\min(f \circ g)^{-1}\{x\}$, then $h:f(A) \to h(f(A))$ is bijective and $h(f(A)) \subseteq I_n$. Thus $f(A)$ is finite.

Thank you for your help!

Answer 1

Anytime you have a surjection from $A$ to $B$, it follows that $\lvert B\rvert\le\lvert A\rvert$. The proof uses the axiom of choice...

See the link.