Ensembles et applications - Institut de Mathématiques de Bordeaux

A={2,3,4}
∀x∈N,(x∈A)
∃x∈N,(x∈A)
∀x, ((x∈N)⇒(x∈A))
∀x∈N,∃y∈N, y ≥x
∃x∈N,∀y∈N, y ≥x
∃x∈Z,∀y∈Z, y ≥x
A={2,3,4}x∈A x ∈Nx= 2z+ 1 x≥4
A
A≥4
E
(E)∃x∈E∀y∈E(x+y=y) (E0)∀x∈E∃y∈E(x+y= 0 )
(E)E
E
(E0)E=NE=Z
P(E)E=∅, E ={0}, E ={0,1,2}
E A B E
A∪(A∩B)A∪(A∪B)
E A B C E (A∩B)∪C6=A∩(B∪C)
A B N
A∩B A 45 B55
A∪B A 2B4
C
A={z∈C| |z−1| ≤ 3}B={z∈C| |z−2i|<2}
RI= [1,3] J= [2,4]
(I∪J)×(I∪J) (I×I)∪(J×J)
f:R→Rf(x) = sin(x)x∈R
f([0, π]) f([−π
4,3π 4])
f−1([−2,1]) f−1({0})
4 (I, J)Rs:I→J
s(x) = sin(x)x∈I
s

s
s
s
X Y
R
X= [0,+∞[Y=RR={(x, y)∈X×Y|x2=y2}
X=RY= [0,+∞[R={(x, y)∈X×Y|x2=y2}
X= [−1,1] Y= [0,+∞[R={(x, y)∈X×Y|x2+y2= 1}
fR R f(x) = x2
[0,1] f−1(f([0,1]))
[−1,1] f¡f−1([−1,1])¢
f([0,1] ∩[−1,0[) f([0,1]) ∩f([−1,0[)
f:E→F
A B E f(A∩B)⊆f(A)∩f(B)
f A B E
f(A∩B) = f(A)∩f(B)
f:N→N
x7→ x+ 1
g:N−→ N
y7−→ n0y=0
y−1
g◦f f ◦g
f: ]0,+∞[→R+f(x) = x+1
x
f
I⊂]0,+∞[J⊂R+f
I J
f:R2→R2f(x, y) = (x−y, −2x+ 2y)
(a, b)∈R2(a, b)∈f(R2) 2a+b= 0
f
f:R2→Rf(x, y) = x−y2
f
f
g:R→R2f◦g= idR
h:R2→R2h(x, y) = (x+y2, y)h
f◦h(x, y)x, y ∈R
n≥1un=
n
P
j=1
j(j+ 1) = 1.2 + 2.3 + ···+n(n+ 1)
n un=n(n+1)(n+2)
3n≥1
13+ 23+ 33+. . . +n3=n2(n+1)2
4n∈N\{0}
fN N \{0}
R R \{0}

3+2i
1−2i
1+i√3
1+i
2πZ1+i√3 1 + i ³(√3 + 1) + (√3−1) i ´20
z∈C\ {0}z+1
z∈R
x
cos(3x) cos(x)
(cos(x))5a1cos(x) + a2cos(2x) + a3cos(3x) + a4cos(4x) + a5cos(5x)a1a2a3
a4a5x
n(1 + i)n
¡n
0¢−¡n 2¢+¡n 4¢−... ¡n 1¢−¡n 3¢+¡n 5¢−...
Cz
(A)z3=−2 (B)z2= 9 + 40 i
(C) (1 −i) z2−(7 + i) z+ 4 + 6i = 0
f:C→Cf(z) = z2z∈C
g:C→Cg◦f= idCg
h:C→Cf◦h= idCh
z∈C\{0}z=reiθr > 0θ∈Rz= exp Z Z = ln r+ iθ
z u =z+1
z
u2+u−1 = 0 z6= 1 z5= 1
u2+u−1 = 1
z2(z4+z3+z2+z+ 1)
cos(2π/5) = (√5−1)/4 cos(π/5) = (√5 + 1)/4
f:C\ {1} → Cf(z) = z2
1−zz∈C\ {1}
f−1({0})f−1({−4})f−1({3i})
u∈Cf−1({u})
f
f:C\ {−i} −→ C
z7−→ z−i
z+i
f
f D ={Z∈C| |Z|<1}
1− |f(z)|2|z+ i|Im(z)
fCΩ = (2,0) r= 2
z∈CI M N R2iziz
z
I M N
[A, B]
IMN

√2 + √3/∈Q
z, z0∈CzRz0⇔ |z|=|z0|.R
1 + i
[0,1] ∩Q,]0,1[∩Q,N,n(−1)n+1
n+ 1 
n∈No.
A B E A4B
A4B=A∪B\A∩B.
A, B, C P(E)A4A=∅A4B=B4A A4(B4C) = (A4B)4C
f:N−→ Zf(2k) = k f(2k+ 1) = −k−1f
I=nf:R−→ R
fbijectiveo(I, ◦)
x−→ ax a
R\Q+×
n Un=nz∈C
zn= 1o
n∈N∗(Un,×)n
p∈N∗(Up,×) (U2p,×)
U=[
n∈N∗
Un(U,×)
K=na+b√2
a, b ∈Qo(K, +,×)
M2(Z) =  a b
c d 
a, b, c, d ∈Z.
M2(Z)

R3R2
F={(x, y, z)∈R3|x+y+z= 0 et 2x−y= 0}
G={(x, y, z)∈R3|x−y= 0}
H={(x, y)∈R2|x−y+ 1 = 0}
J={(x, y)∈R2|x2−y2= 0}
K={(x, y)∈R2|x+ 3y≤4}
L={(t, 4t) ; t∈R}
M={(u+v, u −v) ; u, v ∈R}
R2v= (1,2) w= (−2, m)m∈R
m w v
w v R2
v w
R3v= (1,−2,−5) w= (−2,4, m)m∈R
m w v
w v P
v w
P={(x, y, z)∈R3|ax +by +cz = 0 }
a, b, c
R3v1= (−2,4,1) v2= (1,−2,0) v3= (3, b, −1) b∈R
b v3v1v2
v1v2v3
v2v1v3
R3
v1v2v3
R3
P={(x, y, z)∈R3|x−2y+ 3z= 0 }
v, w P
P v w
R3
R3
u= (3,2,1) v= (4,2,0)
u= (3,1,2) v= (5,1,0) w= (1,1,4)
u= (−2,4,1) v= (1,−2,0) w= (3, m, −1) m