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Definition:

If $F:X \rightarrow Y$ and $A\subseteq X$, then $F(A)=\{y\in Y|y=F(x)\text{ for some x in A}\}$

Proposition

For all subsets A and B of X, $F(A\cap B)=F(A) \cap F(B)$

Let $F$ be a function from set $X$ to set $Y$

We must show that $F(A\cap B)\subseteq F(A)\cap F(B) \text{ and } F(A)\cap F(B) \subseteq F(A\cap B)$

Suppose $y \in F(A \cap B)$. Then

$y= F(x) \text{ for some x in } A \cap B$, which means

$y= F(x) \text{ for some x in } A \text{ and for some x in } B$, which in turn means,

$y= F(x) \text{ for some x in } A$ and $y= F(x) \text{ for some x in } B$ Hence,

$y \in F(A) \text{ and } y\in F(B)$. Thus,

$y\in F(A) \cap F(B)$

Now suppose $y \in F(A) \cap F(B)$. This means that

$y=F(x)\text{ for some x in A}$ and $y=F(x)\text{ for some x in B}$ which means,

$y= F(x) \text{ for some x in } A \text{ and for some x in } B$, which in turn means,

$y= F(x) \text{ for some x in } A \cap B$.

Therefore since $F(A\cap B)\subseteq F(A)\cap F(B) \text{ and } F(A)\cap F(B) \subseteq F(A\cap B)$, $F(A\cap B)= F(A)\cap F(B) $

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  • $\begingroup$ "Now suppose y∈F(A)∩F(B). This means that y=F(x) for some x in A and y=F(x) for some x in B which..." You don't know that it is the same x for A and B though, do you? $\endgroup$ – Rocket Man Aug 28 '15 at 0:00
  • $\begingroup$ The second part is wrong. You used the letter $x$ to mean different things, and it created confusion. Suppose your $F$ is a constant,$c$, and that $A$ and $B$ are disjoint, but non-empty. Then walk through your second part line by line. $\endgroup$ – lulu Aug 28 '15 at 0:01
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It’s the second part of the proof that’s wrong. If $y\in F[A]\cap F[B]$, there is an $x_A\in A$ such that $F(x_A)=y$, and there is an $x_B\in B$ such that $f(x_B)=y$, but there’s no reason to think that $x_A=x_B$. For a simple example, let $X=\{0,1\}$, and define $F:X\to X$ by $F(0)=F(1)=0$. Let $A=\{0\}$ and $B=\{1\}$; then $F[A\cap B]=F[\varnothing]=\varnothing$, but $f[A]=\{0\}=F[B]$.

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