Logique Ensembles - Institut de Mathématiques de Marseille

∃x∈R,∀y∈R, x +y > 0
∀x∈R,∀y∈R, x +y > 0
∃x∈R,∀y∈R, y2> x
0≤x≤y⇒0≤x2≤y2
1 + 2 + . . . +n=n(n+1)
2n∈N
2n≥n2n n0
2n≥n2n∈N
a∀ε > 0a≤ε a ≤0
√2√2p
q
p q
A, B, C E
A∪(B∩C) = (A∪B)∩(A∪C).
A⊂B⇔Bc⊂Ac
(A∩B)c=Ac∪Bc
(A∪B)c=Ac∩Bc
A∪B=B⇔A∩B=A.
{1,2,3}
n2n

A45 B165 A B A∩B A∪B
A= [1,4] B= [3,6] C= [2,5] D={2}
RA∪B A ∩B A \D
R2(A∩B)×A Bc×(A∪C) (A\D)×D
A={x∈R:|x−1|≤|x+ 3|}
B={x∈R:|x2−2x−3|>2x+ 1}
C={x∈R:|x− |x+ 1|| <2}
[t, ∞[ = {x∈R, x ≥t}. A =S
t∈R
[t, ∞[B=T
t∈R
[t, ∞[.
E={a, b, c, d}F={1,2,3,4}f({a, b, c})f−1({1,3})f:
E→F
f(a)=1, f(b)=3, f(c) = 4, f(d)=2
f(a) = f(b) = 2, f(c) = f(d)=3
f(a) = f(b) = f(c) = f(d) = 2
f:E→F A B E
f(A∪B) = f(A)∪f(B)
f(A∩B)⊂f(A)∩f(B)
f(A∩B) = f(A)∩f(B)f
f:E→F C D F
f−1(C∪D) = f−1(C)∪f−1(D)
f−1(C∩D) = f−1(C)∩f−1(D)
f:E→F g :F→G
f g ⇒g◦f
f g ⇒g◦f
g◦f⇒g f
g◦f⇒f g
f:R→Rf(x) = x2
f([−1,1]) Im(f)f−1([0,1]) f−1(] − ∞,0])
f
f:R2→R2,(x, y)7→ (x+y, x −y)
f: [−1,+∞[→[0,+∞[f(x) = (x+ 1)2f
g:R\ {1} → R\ {1}g(x) = x+ 1
x−1g
h:R→Rh(x) = x2+ 2x
I J ˜ h:I→J˜ h(x) = x2+ 2x

f:R→R
∃x∈R:f(x) = 1
∀x∈R:f(x) = 1
∀x, y ∈R:x<y⇒f(x)< f(y)
∃x∈R:x<f(x).
fR R
f
f
f
f
fR R
x∈R, f(x)≤1
f
f
x∈Rx≥0f(x)≤0
x∈Ry∈Rx < y f(x)> f(y)