S1

Question 1

f ◦ g = g ◦ f ⇔ f = g

Answer 1

False

Take for example f (x) = x et g(x) = x² that satisfy (f ◦ g)(x) = x² = (g ◦ f)(x) with f ≠ g.

Question 2

If f ◦ g is injective, then g is injective.

Answer 2

True.

Let x₁, x₂ ∈ ℝ be such that g(x₁) = g(x₂).

So we have f(g(x₁)) = f(g(x₂)). Since f◦g is injective, we conclude that x₁ = x₂.

Question 3

If f ◦ g is injective, then f is injective.

Answer 3

False.

Take for example f(x) = x² et g(x) = e^x defined from ℝ to ℝ. f is not injective but (f ◦ g)(x) = e^2x is injective.

Question 4

If f ◦ g is surjective, then f is surjective.

Answer 4

True.

Let y ∈ ℝ, then there exists x ∈ ℝ such that (f◦g)(x) = y.

So there exists z = g(x) such that f(z) = y, and so f is surjective.

Question 5

Let A, B be two subsets of ℝ, then f(A∩B) = f(A) ∩ f(B).

Answer 5

False.

Take for example A = {0,1}, B = {0,2} and f(0) = a, f(1) = f(2) = b.

Then we have f(A∩B) = {a} ≠ {a,b} = f(A) ∩ f(B).

Question 6

χ_A(x) · χ_B(x) = χ · _C(x) ⇔ A ∩ B = C

Answer 6

True.

From the definition of indicator function we have χ_A(x) · χ_B(x) and χ·C(x).

The proposition follows.

Question 7

ℝ(A ∩ B) = (ℝ\A) ∩ (ℝ\B)

Answer 7

False.

Take for example A = [0,2] et B = [1,3].

We have ℝ(A ∩ B) = ℝ[1,2] and (ℝ\A) ∩ (ℝ\B) = ℝ[0, 3].

Question 8

A×B = B×A ⇔ A = B.

Answer 8

True.

The converse is trivial. We can prove the direct implication by contradiction.

Let A×B = B×A and A ≠ B. Without loss of generality, we assume A ⊄ B and so there exists a ∈ A such that a ∉ B.

Let b ∈ B, then (a,b) ∈ A×B = B×A and so a ∈ B. Contradiction.

Question 9

χ_A×B(x,x) = χ_A∩B(x)

Answer 9

True.

From the definition:

• χ*_A×B(x,x) = 1 if (x,x) ∈ A x B = 1 if x ∈ A ∩ B = χ_A∩B(x) .
• χ*_A×B(x,x) = 0 otherwise