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7/17/2019 DL1 (1)
http://slidepdf.com/reader/full/dl1-1 1/2
MPSI :2015−2016
f : R −→ R
x −→ x
1+|x|
F = f (R)
g : R −→ F
x −→ f (x)
g
g−1
X et Y
f : P (Y ) −→ P (X )
B −→ f −1 (B)
f
f
f f
f : E −→ F S = {X ⊂ E | f −1(f (X )) = X }
A ⊂ E f −1(f (A)) ∈ S
S
∀ (A, B) ∈ S 2, A ∩ B ∈ S et A ∪ B ∈ S
X ∈ S A ⊂ E X ∩ A = ∅ X ∩ f −1(f (A)) = ∅
X Y ∈ S X Y \ X S
ϕ : S −→ P (f (E ))
A −→ f (A)
E = N∗
R
∀ ( p, q ) ∈ E 2, p R q ⇐⇒ ∃ n ∈ N, q = pn
R E
E
{2, 4}
{2, 5}
E
A et B
E
A et B
A∆B = (A \ B) ∪ (B \ A)
A , B et C
E
7/17/2019 DL1 (1)
http://slidepdf.com/reader/full/dl1-1 2/2
A∆B = (A ∪ B) \ (A ∩ B)
A∆ (B∆C ) = (A∆B) ∆C
A ∩ (B∆C ) = (A ∪ B) ∆ (A ∪ C )
F
E
A
E
A∆F = F ∆A = A
A
E
A
E
A∆A = A∆A = F
A E A χA : E −→ {0, 1}
x −→
1 si x ∈ A
0 si x /∈ A
·
A et B ∈ P (E ) χA = χB
A A ∩ B χA et χB
A \ B
A ∪ B
A∆B
A , B et C
E A∆ (B∆C ) = (A∆B) ∆C
A ∩ (B∆C ) = (A ∩ B) ∆ (A ∩ C )
A , B ∈ P (E )
ΦA : P (E ) −→ P (E )
X −→ A∆X
P (E )
ΦA (X ) = B (2)
X ∈ P (E )
E
A∆A et ∅∆X
ΦA (ΦA (X ))
ΦA : P (E ) −→ P (E )
A et B
A et B