Set Theory Question

Question

Prove f(a1∩f^-1(b2)=f(a1)∩b2

Answer 1

To prove, f(A₁ ∩ f^-1(B₂)) = f(A₁) ∩ B₂

we show that,

f(A₁ ∩ f^-1(B₂)) ⊆ f(A₁) ∩ B₂ and

f(A₁) ∩ B₂ ⊆ f(A₁ ∩ f^-1(B₂))

Note that, f^-1(B₂) = {y : f(y) ∈ B₂} ........ (1)

Let, x ∈ f(A₁ ∩ f^-1(B₂))

⇒ there exists y ∈ A₁ ∩ f^-1(B₂) such that f(y) = x

⇒ there exists y such that y ∈ A₁ and y ∈ f^-1(B₂) and f(y) = x

Since y ∈ A₁, then x ∈ f(A₁)

Similarly, since y ∈ f^-1(B₂), we have that x ∈ f(f^-1(B₂)), that is x ∈ B₂

Since x ∈ f(A₁) and x ∈ B₂, we conclude that x ∈ f(A₁) ∩ B₂

This proves that f(A₁ ∩ f^-1(B₂)) ⊆ f(A₁) ∩ B₂

Now for the other part of the proof, let x ∈ f(A₁) ∩ B₂

⇒ x ∈ f(A₁) and x ∈ B₂

⇒ there exists y ∈ A₁ such that f(y) = x and x ∈ B₂

⇒ there exists y ∈ A₁ such that f(y) ∈ B₂ and f(y) = x

⇒ there exists y ∈ A₁ such that y ∈ f^-1(B₂) and f(y) = x [from equation (1)]

⇒ there exists y such that f(y) = x and y ∈ A₁ and y ∈ f^-1(B₂)

⇒ there exists y such that f(y) = x and y ∈ A₁ ∩ f^-1(B₂)

⇒ x ∈ f(A₁ ∩ f^-1(B₂))

This proves that f(A₁) ∩ B₂ ⊆ f(A₁ ∩ f^-1(B₂))

f(A₁ ∩ f^-1(B₂)) = f(A₁) ∩ B₂