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,

[itex]F(A \cap B) \subseteq F(A) \cap F(B)[/itex]

The Attempt at a Solution

Let [itex]y \in F(A \cap B)[/itex]. Then, [itex]\exists x \in A \cap B[/itex], so [itex]\exists x \in A[/itex] and [itex]x \in B[/itex].

Then, [itex]y \in F(A)[/itex] and [itex]y \in F(B)[/itex], so [itex]y \in F(A) \cap F(B)[/itex].

Therefore, [itex]y \in F(A \cap B) \Rightarrow y \in F(A) \cap F(B)[/itex], and hence, [itex]F(A \cap B) \subseteq F(A) \cap F(B)[/itex].

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,

[itex]F(A \cap B) \subseteq F(A) \cap F(B)[/itex]

The Attempt at a Solution

Let [itex]y \in F(A \cap B)[/itex]. Then, [itex]\exists x \in A \cap B[/itex], so [itex]\exists x \in A[/itex] and [itex]x \in B[/itex].

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

Then, [itex]y \in F(A)[/itex] and [itex]y \in F(B)[/itex], so [itex]y \in F(A) \cap F(B)[/itex].

Therefore, [itex]y \in F(A \cap B) \Rightarrow y \in F(A) \cap F(B)[/itex], and hence, [itex]F(A \cap B) \subseteq F(A) \cap F(B)[/itex].

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 "[itex]\subseteq[/itex]" rather than just "[itex]\subset[/itex]".

 

[SithsNGiggles]

[itex]\exists x \in A\cap B[/itex] 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 "[itex]\subseteq[/itex]" rather than just "[itex]\subset[/itex]".

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.