So, $A$ and $B$ are non empty finite sets and there is a function $f:A \to B$ and I need to prove that $|A| \geqslant |f(A)| $. So, $|A|$ is the cardinality of set $A$ and $ |f(A)| $ is the cardinality of the co-domain of the function. So, here is my thinking on this. Since $f$ is a function, each element in $A$ is mapped to some element in $B$. In the worst case, all elements of $A$ are mapped to a single element of $B$. And since $A$ is non-empty, there is at least one element in $A$. So, if there is only one element in $A$, then we have $|A| = |f(A)| $. Another case is when $|A| > 1$ and $ |f(A)| = 1$. Here we would have $|A| > |f(A)| $. And another case is when the function is one to one. So, every element of $A$ must be mapped to a different element of $B$. So, we must have $|A| = |f(A)|$. So, in any case, we have $|A| \geqslant |f(A)| $. Now, would this be considered a valid proof ? I don't know if this proof is without any loopholes.
Thanks
Best Answer
For each $y\in f(A)$ choose an $x\in A$ with $f(x)=y$. In this way $|f(A)|$ different points $x\in A$ are chosen, forming a subset $A'\subset A$. We then have $$|A|\geq|A'|=|f(A)|\ .$$