Is this proof correct for : Does $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$?

Solution 1:

Your gut is right. Something is indeed very wrong: the statement that you’re trying to prove is false. Here’s a familiar counterexample: let $F:\Bbb R\to\Bbb R:x\mapsto x^2$, let $A$ be the set of negative real numbers, and let $B$ be the set of positive real numbers. Then $F[A]=F[B]=B$, so $F[A]\cap F[B]=B$, but $A\cap B=\varnothing$, so $F[A\cap B]=\varnothing$. Clearly $B\nsubseteq\varnothing$!

The problem with your reasoning, as you can see from the counterexample, is that although $x\in F[A]$ guarantees that $x=F(a)$ for some $a\in A$, and $x\in F[B]$ guarantees that $x=F(b)$ for some $b\in B$, you have no guarantee that $a=b$. There may be nothing at all in $A\cap B$.

Solution 2:

The third line is mistaken. You only know that there exists an $x$ in $A$ such that $F(x)=y$, and you know there is a $z\in B$ such that $F(z)=y$.

It is extremely easy to find a counterexample: just draw two sets $A$, $B$ that are disjoint, and map an $a\in A$ and a $b\in B$ to a single point. Then you have that $y\in F(A)\cap F(B)$, but $F(A\cap B)=\emptyset$.

Solution 3:

This is wrong. Consider $A=\{1,2\}$ and $B=\{3,4\}$ and we can define $F(1)=F(3)=1,a$ not equal $c$ and this $A\cap B= \emptyset$ while $F(A)\cap F(B)$ doesn't equal to $\emptyset$. This is a counter example to your statement

Solution 4:

Here, $f: A \rightarrow B$ is in green and $\{S_i\} = S_i$ for $S = A, B$ and $i = 1, 2.$
This picture proves that $ f(A_1) \cap f(A_2) \not\subseteq f(A_1 \cap A_2)$. Incidentally, the same picture works for Proving $f(C) \setminus f(D) \subseteq f(C \setminus D)$ and disproving equality.