Is this proof correct? To prove $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$.

Question

Is this proof correct? To prove $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$.

Let any number $y\in F(A)\cap F(B)$. We want to show $y\in F(A\cap B).$

Therefore, $y\in F(A)$ and $y\in F(B)$, by definition of intersection.

By definition of inverse, $y=F(x)$ for some $x\in A$ and $x\in B$

And so, $y=F(x)$ for some $x\in A\cap B$

And therefore, $y\in F(A\cap B)$

I have a gut feeling deep down that something is wrong. Can anyone help me pinpoint the mistake? I am not sure why am I having so much problems with functions. Am I not thinking in the right direction?

Answer 1

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 $F(A\cap B)=\emptyset$.

Answer 2

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 $A\cap B$.

Answer 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