Real Analysis beginner proof

[Darya]

I'm self-learning RA and I've got a question

Assume that I have proved that for a function g: R -> R is always true that g(A ∩ B) ⊆ g(A) ∩ g(B) for all sets A, B ⊆ R.
Can I also prove that g(A ∩ B) ⊇ g(A) ∩ g(B)? If so, where should I start? ( I'm a complete noob at proofs yet)

Thanks!

 

[pka]

Assume that I have proved that for a function g: R -> R is always true that g(A ∩ B) ⊆ g(A) ∩ g(B) for all sets A, B ⊆ R. Can I also prove that g(A ∩ B) ⊇ g(A) ∩ g(B)? If so, where should I start? ( I'm a complete noob at proofs yet)

I take it that you have proved that the image of an intersection is a subset of the intersection of the images.
Now you want to see if you can prove that \(g(A)\cap\ g(B)\subseteq g(A\cap B)~?\)
So the standard proof would be if \(t\in g(A)\cap\ g(B)\) then \(t\in g(A)\wedge t\in g(B)\). RIGHT?
That means \((\exists a_t\in A)[g(a_t)=t]\wedge (\exists b_t\in B)[g(b_t)=t]\)RIGHT?
Can you conclude what you need?

 

[Darya]

I take it that you have proved that the image of an intersection is a subset of the intersection of the images.
Now you want to see if you can prove that \(g(A)\cap\ g(B)\subseteq g(A\cap B)~?\)
So the standard proof would be if \(t\in g(A)\cap\ g(B)\) then \(t\in g(A)\wedge t\in g(B)\). RIGHT?
That means \((\exists a_t\in A)[g(a_t)=t]\wedge (\exists b_t\in B)[g(b_t)=t]\)RIGHT?
Can you conclude what you need?

Thanks for your reply! But wait, if we also prove that, it'll mean g(A∪B)= g(A) ∩ g(B), right? But if we take for example g(x)={x²:x€R} it's not true. We can take set A={-1,0}, B={0,1}. Then g(A∪B)=0, however g(A) ∩ g(B)={0,1}. Am I missing something?

 

[pka]

Thanks for your reply! But wait, if we also prove that, it'll mean g(A∪B)= g(A) ∩ g(B), right? But if we take for example g(x)={x²:x€R} it's not true. We can take set A={-1,0}, B={0,1}. Then g(A∪B)=0, however g(A) ∩ g(B)={0,1}. Am I missing something?

Suppose that \(A=\mathbb{Z}^+~\&~B=\mathbb{Z}^-\). If \(g(x)=|x|\) can you prove what you claim?

 

[Darya]

Suppose that \(A=\mathbb{Z}^+~\&~B=\mathbb{Z}^-\). If \(g(x)=|x|\) can you prove what you claim?

I reformed it a little.

If C ⊆ Z+, D⊆ Z-, g(x)={|x|:x€Z}, ∃x € C => ∃y€ D: y=-x.
Thus, ∃A={x}, ∃B={y:y=-x}
g(A)=x, g(B)=x. If t€g(A) and t€g(B) => t€ (g(A) ∩ g(B)). Thus, g(A) ∩ g(B)=x.

Similarly, if t€A and t€B => t€A∩B.
With that said, A∩B=∅.
g(A∩B)=g(∅)=∅ => g(A)∩g(D) is not subset of g(A∩B).

Correct?