feuille 1 - Institut de Mathématiques de Toulouse

YjY YjY
!
n|a1+a2+···+an|≤|a1|+|a2|+···+|an|
a1, a2, . . . , an∈R
A, B RA+B:= {a+b, a ∈A, b ∈B},−A:= {−a, a ∈
A}inf(−A) = −sup(A) inf(A+B) = inf A+ inf B
A⊂R
f E F A E B
F
f(f−1(B)) ⊂B
A⊂f−1(f(A))
n∈N?En:= {k+k
n, k ∈N?}
Eninf En:= inf{k+k
n,1≤k≤n}
n∈N?inf En≥√4n
!S⊂Rsup S6∈ Sx < sup S
Sxsup S
!f, g :X→RXsupx∈Xf(x)+
g(x)≤supx∈Xf(x) + supx∈Xg(x)f(x) = sin(x), g(x) = cos(x), X =R
sup(A+B) = sup A+ sup B
x∈Rn n ≤x≤n+1
x E(x)
•x≤y=⇒E(x)≤E(y)
•E(x) + E(y)≤E(x+y)≤E(x) + E(y)+1
• ∀n∈N?, x ∈R:EE(nx)
n=E(x)
• ∀n∈N?:√n+ 1 −√n < 1/2√n < √n−√n−1
1
21
√2+1 √3+··· +1 √10000 

!•A⊂Rsup(A)>0a∈A
a > 0
•A, B RA∪B
sup(A∪B) inf(A∪B)
•B⊂A A B sup A, sup B, inf A, inf B
•f0∈A f(0) = 0
•f:A→R
f=g+h g h
•f(x) = cos(x2)g(x) = sin(x2)
•fRf(x) = 1 x∈Qf(x) = 0 6∈ Q
f
!•E, F, G f :E→F, g :F→G g ◦f
f g ◦f g
•
•
f:E→F
•(∀X∈P(E) : f−1(f(X)) = X)⇐⇒ (f)
•(∀Y∈P(F) : f(f−1(Y)) = Y)⇐⇒ (f)
!f:X→Y
•f
•A, B X f(A∩B) = f(A)∩f(B)
f:E→F f A
P(E)f(A) = f(A)A A
!EP(E)A B
E
f:P(E)→P(A)×P(B)
X7→ (X∩A, X ∩B).
f A ∪B=E
f A ∩B=∅
A B f
!f:N2→N∗(n, p)7→ 2n(2p+ 1) f
N2N

f:X→Y
•f
•A, B X f(A∩B) = f(A)∩f(B)
1 =⇒2y=f(A∩B)x∈A∩B
y=f(x)y∈f(A)y=f(x)x∈A y ∈f(B)
y∈f(A)∩f(B)y∈f(A)∩f(B)a∈A y =f(a)b∈B
y=f(b)f a =b a ∈A∩B y ∈f(A∩B)
2 =⇒1a b f(a) = f(b) = y A ={a}B={b}f(A) = f(B) = {y}
f(A∩B) = {y}A∩B6=∅a=b
f:E→F f A P(E)f(A) = f(A)
A A
f A P(E)
x f(A)x=f(y)y∈A x ∈f(A)x=f(z)
z∈A f(y) = f(z)f y =z y A z
x /∈f(A)f(A)⊂f(A)
x∈f(A)f y E
x=f(y)y A x f(A)y
A x ∈f(A)
AP(E)f(A) = f(A)
f x, y f(x) = f(y)x6=y
A={x}y∈A f(y)∈f(A)⊂f(A)f(A) = {f(x)}f(y)6=f(x)
f A =E f(E) = f(E)f(E) = f(∅) = ∅
f(E) = ∅f(E) = F f 
EP(E)A B E
f:P(E)→ P(A)× P(B)
X7→ (X∩A, X ∩B).
f A ∪B=E
f A ∩B=∅
A B f
A∪B6=E x ∈E\(A∪B)X={x}
f(X)=(X∩A, X ∩B)=(∅,∅)x A B f(∅)=(∅,∅)
f(X) = f(∅)X6=∅f
X⊂E A ∪B=E
X=X∩E=X∩(A∪B) = (X∩A)∪(X∩B).
X, X0⊂E f(X) = f(X0)X∩A=X∩A0X∩B=X∩B0
X= (X∩A)∪(X∩B) = (X0∩A)∪(X0∩B) = X0.
f
f x ∈A X ⊂E f(X) = ({x},∅)
X∩B=∅x∈X∩A x ∈X x /∈B A ∩B=∅
A∩B=∅A0⊂A B0⊂B X =A0∪B0A∩B=∅
X∩A=A0X∩B=B0f(X) = (A0, B0)f
f A∪B=E A ∩B=∅(A, B)
E(A0, B0)7→ A0∪B0
f:N2→N∗(n, p)7→ 2n(2p+ 1) fN2
N

fN∗f(n, p)
(m, q)N2f(n, p) = f(m, q) 2n(2p+ 1) = 2m(2q+ 1).
n≥m
2n−m(2p+ 1) = 2q+ 1.
n6=m n =m
2p+ 1 = 2q+ 1 p=q(n, p)=(m, q)f f
l∈N∗l
l= 2npα2
2. . . pαr
r,
pii≥2pα2
2. . . pαr
r
2p+ 1 p∈Nl= 2n(2p+ 1) = f(n, p)f
N2N N∗Nn7→ n−1
g: (n, p)7→ 2n(2p+ 1) −1N2N