Prove or Counter example.For all nonempty sets $A$ and $B$ and for all functions $F: A \to B$, $F(A-B) = F(A) - F(B)$

Question

Prove or counter-example. For all nonempty sets $A$ and $B$ and for all functions $F$, $F(A-B) = F(A) - F(B)$; if not, what else does $F$ need to have in order to make the equality hold?

I am pretty lost on this question. I don't feel like its right since it would be a pretty basic proof but I can't find a counterexample.

Answer 1

We prove the following: $F(A)−F(B)\subset F(A-B) $

First we prove: $F(A-B)\cup F(B)=F(A)\cup F(B)$

\begin{align} F(A-B)\cup F(B)&=F((A-B)\cup B) \\ &=F((A\cap B^c)\cup B) \\ &=F((A\cup B) \cap (B^c\cup B)) \\ &=F(A\cup B) \\ &=F(A)\cup F(B) \end{align}

Then \begin{align} F(A)−F(B)&=(F(A)-F(B)) \cup \varnothing \\ &=(F(A)-F(B)) \cup (F(B)-F(B)) \\ &=(F(A)\cup F(B))-F(B) \\ &=(F(A-B)\cup F(B))-F(B) \\ &=(F(A-B)-F(B))\cup (F(B)-F(B)) \\ &=(F(A-B)-F(B))\cup \varnothing \\ &=F(A-B)-F(B) \\ &\subset F(A-B) \end{align}

So $F(A)−F(B)\subset F(A-B) $

A counterexample for $F(A-B)\subset F(A)-F(B) $ is as:

let $F(x)=\tan(x), A=[-\dfrac{\pi}{2},\dfrac{\pi}{2}], B=(\dfrac{\pi}{2},\pi)$, then

$A-B=\varnothing, F(A-B)=(-\infty,\infty)$, but

$F(A)=(-\infty,\infty), F(B)=(-\infty,0)$, and

$F(A-B)\not\subset F(A)-F(B)=[0,\infty)$

Answer 2

Minimal counterexample: $X=\{1,2\}$, $Y=\{0\}$, $f\colon X\to Y$ is the only possible function. If $A=\{1,2\}$ and $B=\{2\}$, then...

Answer 3

Let $F:X\to Y$ be constant function: $\forall x\in X, F(x)=c$, where $c$ is a constant element of $Y$. Let $A$ and $B$ be subsets of $X$, such that $A-B$ is not empty. Then $$F(A)−F(B)= \varnothing,$$ while, $$F(A-B)=\{c\}$$