Proving image of intersection?

[SithsNGiggles]

Homework Statement

Let F be a relation from X to Y and let A and B be subsets of X. Then,

$F(A \cap B) \subseteq F(A) \cap F(B)$

The Attempt at a Solution

Let $y \in F(A \cap B)$. Then, $\exists x \in A \cap B$, so $\exists x \in A$ and $x \in B$.

Then, $y \in F(A)$ and $y \in F(B)$, so $y \in F(A) \cap F(B)$.

Therefore, $y \in F(A \cap B) \Rightarrow y \in F(A) \cap F(B)$, and hence, $F(A \cap B) \subseteq F(A) \cap F(B)$.

I'm having trouble showing that the right side is not a subset of the left. Thanks for any help.

 

[HallsofIvy]

Homework Statement

Let F be a relation from X to Y and let A and B be subsets of X. Then,

$F(A \cap B) \subseteq F(A) \cap F(B)$

The Attempt at a Solution

Let $y \in F(A \cap B)$. Then, $\exists x \in A \cap B$, so $\exists x \in A$ and $x \in B$.

$\exists x \in A\cap B$ such that F(x)= y. You might want to say that!

Then, $y \in F(A)$ and $y \in F(B)$, so $y \in F(A) \cap F(B)$.

Therefore, $y \in F(A \cap B) \Rightarrow y \in F(A) \cap F(B)$, and hence, $F(A \cap B) \subseteq F(A) \cap F(B)$.

I'm having trouble showing that the right side is not a subset of the left. Thanks for any help.

That's because it may not be true! That's the reason for the "$\subseteq$" rather than just "$\subset$".

 

[SithsNGiggles]

$\exists x \in A\cap B$ such that F(x)= y. You might want to say that!

This is actually the second part of the problem I'm on. The previous one was about the image of a union being equal to the union of the images. I mentioned your suggestion in that part.

That's because it may not be true! That's the reason for the "$\subseteq$" rather than just "$\subset$".

I understand that. I have to show that it's NOT a subset of the left hand side. I just don't know how to do that using the kind of logic I used in proving the left side was a subset of the right.

 

[HallsofIvy]

And I will say again that you can't prove that- it is not true- unless you mean proper subset.