L3B Topologie Contrôle continu du 13 décembre 2013 13h30

(E, ||.||)
A E
K E
A B E A ∪B
S2R3
S2N= (0,0,1)
(C,|.|)C
|z|=px2+y2z=x+iy x, y ∈R
r≥0Ar⊂CAr=B(−1, r)∪¯ B(1,2r)
ArB(−1, r)¯ B(1,2r)
Arr= 1/2r= 1 r= 3
r≥0Ar
¯
Arr≥0Ar
r δ(r) = supx,y∈Ar|x−y|
r≥0Ar
l∞(un)n∈N
|| · ||∞||(un)||∞= supn∈N|un|
|| · ||∞l∞
f:l∞→l∞,(un)7→ (vn), vn=un+1 −un.
f|||f|||
(E, ||.||)A B
E
A∩B A ∪B
A B
A∩B
b∈A∩B
a∈A γ : [0,1] →A∪B γ(0) = a γ(1) = b
t0= sup{t∈[0,1]|γ([0, t]) ⊂A}γ(t0)∈A∩B
A
A∩B A ∪B A
B

(E, ||.||)
A E (an)A
ϕ:N→N(aϕ(n))
Oi, i ∈I Oi⊂E
A⊂ ∪i∈IOiJ⊂I
A⊂ ∪i∈JOi
K E (an)K l ∈E K
(aϕ(n))ϕ:N→Na∈K
(an)l(an) (an)a∈K K
K E B(k, 1), k ∈K
K K
J⊂K K ⊂ ∪k∈JB(k, 1)
∀x∈K, ||x|| ≤ maxk∈J(||k|| + 1)
A B E A ∪B Oi, i ∈I
A∪B⊂ ∪i∈IOiOiA B I1⊂I I2⊂I A ⊂ ∪i∈I1Oi
B⊂ ∪i∈I2OiJ=I1∪I2⊂I A ∪B⊂∪i∈JOiA∪B
A∪B
(R3,||.||)
S2={v∈R3tel que ||v|| = 1}=f−1({1})f(v) = ||v|| f
S2{1}Rf S2⊂¯ B(0,1) S2
S2\ {N}N= (0,0,1) vn=
(sin(1/n),0,(cos(1/n)) ∈S2\ {N}N= (0,0,1) 6∈ S2\ {N}S2\ {N}
(C,|.|)C|z|
r≥0Ar⊂CAr=B(−1, r)∪¯ B(1,2r)
ArB(−1, r)¯ B(1,2r)
Arr= 1/2,2/3,1 3
r≥0Ar=B(−1, r)∪B(1,2r)
z ∂ ¯ B(1,2r)
B(z, ε)Ar
r≥0¯ Ar=¯ B(−1, r)∪¯ B(1,2r)
(C,|.|)Ar
¯
Ar=¯ B(−1, r)∪¯ B(1,2r) = B(−1, r)∪¯ B(1,2r)
¯
B(−1, r)⊂B(−1, r)∪¯ B(1,2r)∂B(−1, r)⊂¯ B(1,2r)−1−r≥1−2r⇐⇒ r≥2

B(r/2,1+3r/2)
B(1,2r)
B(-1,r)
b
a
Ar
δ(r) = supx,y∈Ar|x−y|= 4r, r ≥2,2 + 3r, r ≤2
δ(r) = diamAr= diam ¯ Ar
diamAr≤diam ¯ ArAr⊂¯ Ara, b ∈¯ Ardiam ¯ Ar=|a−b|¯ Ar
(C,|.|) (x, y)→ |x−y|anbnAr
a b diamAr≥ |an−bn| → |a−b|= diam ¯ ArdiamAr≥diam ¯ Ar
r≥2¯ B(−1, r)⊂¯ B(1,2r)¯ Ar=¯ B(1,2r) diam ¯ Ar= 4r
r≤2a=−1−r b = 1 + 2r¯ Ar=¯ B(−1, r)∪¯ B(1,2r)⊂¯ B(r/2,1+3r/2)
ab a b
¯
Arab ¯ Ardiam ¯ Ar= 2 + 3r
H−={(x, y)∈R2tel que x < −1/3}H+={(x, y)∈R2tel que x > 1/3}
r < 2/3, B(−1, r)⊂H−B(1,2r)⊂H+Ar
r= 0 B(−1, r) = ∅A0={1}
r≥2/3ArS= [−1,1] B(−1, r)¯ B(1,2r)
Ar= ( ¯ B(−1, r)∪S)∪¯ B(1,2r)
f:l∞→l∞,(un)7→ (vn), vn=un+1 −un.
f||f((un))||∞= supn∈N|un+1 −un| ≤ supn∈N(|un+1|+|un|)≤2||(un)||∞f
|||f||| ≤ 2un= (−1)nvn= 2(−1)n+1 ||(un)||∞= 1 ||f((un))||∞= 2
|||f||| = 2
A B 6=∅A∩B A ∪B
A∩B A ∪B A B A ∪B
b∈A∩B
a∈A A ∪B γ : [0,1] →A∪B γ(0) = a
γ(1) = b
t0= sup{t∈[0,1]|γ([0, t]) ⊂A}τnγ([0, τn]) ⊂Alim τn=t0
γlim γ(τn) = γ(t0)A γ(t0)∈A
γ(t0)∈B γ(t0)∈Bcγ t0
ε > 0, γ(t)∈Bcγ(t)∈A γ(t)∈A∪B t ∈]t0−ε, t0+ε[γ([0, t0+ε/2]) ⊂A
t0
a∈A c ∈A∩B A ∩B
b a ∈A A b a, a0A A
A a b b a0A
A∩B A ∪B A B
A F ⊂A F 0⊂A A =F∪F0
A∩B=G∪G0G=F∩B G0=F0∩B A ∩B G =∅
G0=∅G=∅G0=A∩B G B
H=F0∪B A ∪B=F∪H F H A ∪B
A