2012-13 L3 MAF Topologie & analyse hilbertienne Feuille 1

R R G
(R, +) G = aZ a
G R R G
G + = {g ∈ G , g > 0} m = inf G +
m > 0 m ∈ G + G = mZ
m = 0 m ∈ G + G R
a, b b = 0 a
b /∈ Q Ga,b := aZ + bZ
R R
a /∈ Q {cos(na) , n ∈ Z} [−1, +1]
π
Ga,2π
R 2 R 2
A = (] − ∞, −1] × {0}) ∪ ([−1, 1[×[−1, 1[) ∩ (R ∗ × R) B = Q × (R \ Q)
C = {(x, y) ∈ R 2 x 2 + y 2 ≤ 2, (x − 1) 2 + y 2 > 1} D = {(x, y) ∈ R 2 x 2 − sin y ≤ 4}
E = {(x, y) ∈ [0, 1] 2 cos x > 0} k ∈ R ∗ + Fk = ∪n∈N ∗Hk,n Fk,n
(1/n, 1/n) k/n
A X
A X
(X, d) Y
X f : Y → R k ∀x, y ∈ Y |f(x) − f(y)| ≤
k d(x, y) x ∈ X y ∈ Y fy(x) = f(y) + k d(x, y)
x ∈ X {fy(x)|y ∈ Y } f(z) − k d(z, x)
z ∈ Y g(x)
x ∈ Y g(x) = fx(x)
g : X → R k
f X
Y X
E (Ai)i∈I
E ∪i∈I Ai (∪i∈IAi) E = R 2 I = N ∗ Ai = {(1/i, 1/j) | j ∈
N ∗ }
(∩i∈IAi) ⊂ ∩i∈I Ai
A
A (A) A
A
A A ∩ (A) = ∅
A (A) ⊂ A
A (A) = ∅
A ⊂ (A) ( A) ⊂ (A) R
A B
(A ∪ B) ⊂ (A) ∪ (B)
R
Ā ∩ ¯ B = ∅ ⇒ (A ∪ B) = (A) ∪ (B)