f ◦ g = g ◦ f ⇔ f = g
False
Take for example f (x) = x et g(x) = x2 that satisfy (f ◦ g)(x) = x2 = (g ◦ f)(x) with f ≠ g.
If f ◦ g is injective, then g is injective.
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.
If f ◦ g is injective, then f is injective.
False.
Take for example f(x) = x2 et g(x) = ex defined from ℝ to ℝ. f is not injective but (f ◦ g)(x) = e2x is injective.
If f ◦ g is surjective, then f is surjective.
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.
Let A, B be two subsets of ℝ, then f(A∩B) = f(A) ∩ f(B).
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).
χA(x) · χB(x) = χ · C(x) ⇔ A ∩ B = C
True.
From the definition of indicator function we have χA(x) · χB(x) and χ·C(x).
The proposition follows.
ℝ\(A ∩ B) = (ℝ\A) ∩ (ℝ\B)
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].
A×B = B×A ⇔ A = B.
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.
χA×B(x,x) = χA∩B(x)
True.
From the definition:
χA×B(x,x) = 1 if (x,x) ∈ A x B = 1 if x ∈ A ∩ B = χA∩B(x) .
χA×B(x,x) = 0 otherwise
