- #1

fatineouahbi

- 10

- 0

2.

**Prove f(A) ⋂ f(B) ⊂ f(A⋂B) ⟺ f is an injection**

I've solved the first question , as for the second I started with f(A) ⋂ f(B) ⊂ f(A⋂B) ⇒ f is an injection this way :

Let's suppose f(a) = f(b) = p

If we consider A = {a} and B = {b} then f(A) = f(B) = p

then f(A) ⋂ f(B) = p

then f(A⋂B) = p (because f(A) ⋂ f(B) ⊂ f(A⋂B) from the supposition and f(A⋂B) ⊂ f(A) ⋂ f(B) from the first question)

then A⋂B ≠ ∅

then a=b

then f is an injection .

But I don't know how to solve "f is an injection ⇒ f(A) ⋂ f(B) ⊂ f(A⋂B)"