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2012-13 L3 MAF Topologie & analyse hilbertienne Feuille 1

2012-13 L3 MAF Topologie & analyse hilbertienne Feuille 1

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X <br />

(Ai)i∈I (Bj)j∈J X<br />

(∪i∈IAi) ∩ (∪j∈JBj) = ∪(i,j)∈I×J(Ai ∩ Bj) (∩i∈IAi) ∪ (∩j∈JBj) = ∩(i,j)∈I×J(Ai ∪ Bj)<br />

((Aj)j∈Ji )i∈I X<br />

∪i∈I(∪j∈Ji Aj) = ∪j∈∪i∈IJi Aj ∩i∈I(∩j∈Ji Aj) = ∩j∈∩i∈IJi Aj<br />

(Ai)i∈I (Bi)i∈I X<br />

( <br />

Ai) ∩ ( <br />

Bi) = <br />

(Ai ∩ Bi)<br />

i∈I<br />

i∈I<br />

i∈I<br />

X, Y f : X → Y <br />

A X f(A) = {f(x) x ∈ A} = {y ∈ B | ∃x ∈ A, y = f(x)}<br />

B Y f −1 (B) = {x ∈ A | f(x) ∈ B}<br />

∀A ∈ P(X), A ⊂ f −1 (f(A)) <br />

f injective ⇔ ∀A ∈ P(X), f −1 (f(A)) = A.<br />

<br />

∀B ∈ P(Y ), f(f −1 (B)) ⊂ B <br />

f surjective ⇔ ∀B ∈ P(Y ), f(f −1 (B)) = B.<br />

<br />

(Ai)i∈I X<br />

f(∪i∈IAi) = ∪i∈If(Ai) f(∩i∈IAi) ⊂ ∩i∈If(Ai) f <br />

<br />

(Bi)i∈I Y <br />

f −1 (∪i∈IBi) = ∪i∈If −1 (Bi) f −1 (∩i∈IBi) = ∩i∈If −1 (Bi)<br />

A ∈ P(X) B ∈ P(Y ) f −1 (B c ) (f −1 (B)) c f(A c ) (f(A)) c<br />

A, B A × B <br />

<br />

<br />

<br />

z <br />

z<br />

{0, 1} N Y X <br />

X Y Y N Y <br />

(E, d) δ(A) = sup x,y∈A d(x, y) <br />

A E δ(∅) = 0<br />

δ(A) = 0 A E δ( Ā) = δ(A)<br />

R Ai Ai (Ai) <br />

A1 = [−3, 1[∪]1, 2] ∪ {3, π} A2 = Z A3 = {(−1) p + 1/2p | p ∈ Z} A4 = Q A5 =] − ∞, 3/2] ∩ Q


R R G <br />

(R, +) G = aZ a <br />

G R R G<br />

G + = {g ∈ G , g > 0} m = inf G +<br />

m > 0 m ∈ G + G = mZ<br />

m = 0 m ∈ G + G R<br />

a, b b = 0 a<br />

b /∈ Q Ga,b := aZ + bZ <br />

R R<br />

a /∈ Q {cos(na) , n ∈ Z} [−1, +1]<br />

π<br />

Ga,2π<br />

R 2 R 2 <br />

<br />

A = (] − ∞, −1] × {0}) ∪ ([−1, 1[×[−1, 1[) ∩ (R ∗ × R) B = Q × (R \ Q)<br />

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}<br />

E = {(x, y) ∈ [0, 1] 2 cos x > 0} k ∈ R ∗ + Fk = ∪n∈N ∗Hk,n Fk,n <br />

(1/n, 1/n) k/n<br />

A X <br />

A X<br />

(X, d) Y <br />

X f : Y → R k ∀x, y ∈ Y |f(x) − f(y)| ≤<br />

k d(x, y) x ∈ X y ∈ Y fy(x) = f(y) + k d(x, y)<br />

x ∈ X {fy(x)|y ∈ Y } f(z) − k d(z, x)<br />

z ∈ Y g(x) <br />

x ∈ Y g(x) = fx(x)<br />

g : X → R k<br />

f X <br />

Y X <br />

E (Ai)i∈I <br />

E ∪i∈I Ai (∪i∈IAi) E = R 2 I = N ∗ Ai = {(1/i, 1/j) | j ∈<br />

N ∗ } <br />

(∩i∈IAi) ⊂ ∩i∈I Ai <br />

A <br />

A (A) A<br />

A <br />

A A ∩ (A) = ∅<br />

A (A) ⊂ A<br />

A (A) = ∅<br />

A ⊂ (A) ( A) ⊂ (A) R <br />

<br />

A B <br />

(A ∪ B) ⊂ (A) ∪ (B)<br />

R <br />

Ā ∩ ¯ B = ∅ ⇒ (A ∪ B) = (A) ∪ (B)

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