Examen de Topologie

⇒A⊂B(x, r)x∈X r > 0a, a0∈A d(a, a0)≤
d(a, x) + d(x, a0)≤2rdiam(A)≤2r
⇒x∈X r > 0A⊂B(x, r)a∈A
r= diam(A)+d(a, x)a0∈A d(a0, x)≤d(a0, a)+d(a, x)≤
diam(A) + d(a, x) = r
⇒
diam(A)≤diam(A)ε > 0
b, b0∈A a, a0∈A a ∈B(b, ε)a0∈B(b0, ε)
d(b, b0)≤d(b, a) + d(a, a0) + d(a0, b0)≤ε+ diam(A) + ε ε > 0
d(b, b0)≤diam(A) diam(A)≤diam(A)
d(A, B) = infa∈A,b∈Bd(a, b)A B
c, c0∈A∪B d(c, c0)≤diam(A) + d(A, B) + diam(B)c, c0∈A
d(c, c0)≤diam(A)c, c0∈B
c∈A c0∈B ε > 0
d(A, B)a∈A b ∈B d(a, b)≤d(A, B) + ε
d(c, c0)≤d(c, a) + d(a, b) + d(b, c0)≤diam(A) + d(A, B) + ε+ diam(B)
ε d(c, c0)≤diam(A) + d(A, B) + diam(B)
A∪Bdiam(A∪B)≤diam(A) + d(A, B) + diam(B)
d1((un)n∈N,(vn)n∈N) = X
n
|vn−un| ≤ X
n
|vn|+X
n
|un|
(un),(vn)
(wn)l1
d1((un),(wn)) = X
n
|wn−un|=X
n
|wn−vn+vn−un| ≤ X
n
|wn−vn|+X
n
|vn−un|
d1((un),(wn)) ≤d1((un),(vn)) + d1((vn),(wn))
(δk)k
δk,n = 0 k6=n δk,k = 1
k d1(δk,0) = 1
(δk)k∈N(δk)k∈N
k1, k2d1(δk1, δk2)
(δk)k∈Nd1(δk1, δk2) = 2 k16=k2

a1,· · · , anPa2
i≤(Pai)2
(un)n∈N∈l1
(d2((un)n∈N,0))2≤(d1((un)n∈N,0))2
k > 0a > 0 (uk,n)n∈N
1
kan= 1,· · · , k n > k d1((uk,n)n∈N,0) = k1−a
d2((uk,n)n∈N,0) = k0.5−a0.5<a<1 (uk,n)n∈Nd2
d1
d1d2l1
XT
TX
X∅ T
T
T
f:X→Y
x∈X V f(x)U x f(U)⊂V
f X X
f:X→Y V ⊂Y
x∈f−1(V)V f(x)U
x f(U)⊂V x ∈U⊂f−1(V)f−1(V)
f−1(V)
f Y
X x ∈X V ⊂Y f(x)W
f(x)∈W⊂V U =f−1(W)x f(U)⊂V
f x
R2
k > 0u2k= 1/k u2k+1 =k
X
X= [0,1] ∪[2,3] ⊂R
A= [0,1]
f:Z→Z Z
Z
X=RA=] − ∞,0[ X=R2A