Do I need axiom of choice in this question?

Question

If $X$ is a finite set and $Y$ an infinite one, I'm trying to prove there is a injective function $f:X\to Y$.

My solution: Let $X=\{x_1,\ldots,x_n\}$, for each $x_i$ choose $y_1,\ldots,y_n$ such that $f(x_1)=y_1,\ldots,f(x_n)=y_n$.

I would like to know if I need theorem of choice to choose these $y_i$.

Thanks

Answer 1

No you don't. $Y$ being infinite it is non empty and therefore there is one element in $Y$ say $x_1$. $Y-\{x_1\}$ is not empty otherwise $Y$ would be finite so we can pick $x_2$ etc. You notice that this could work as well if $X$ was countable! So the axiom of choice is not needed either. As a rule of thumb in most cases we only need it for uncountable matters

Answer 2

The axiom of choice is not needed.

You are using finite choice, which is provable in $\sf ZF$. The proof is by induction here, and each step requires choosing one element.

Note that the induction does not prove that there is a countably infinite subset. Just finite subsets of arbitrary size.