Corrigé de l`Examen Partiel de Topologie 2010-2011

(X, O)
[0,1[ R
(X, O)Y X
Y X Y [0,1[
[0,2] R
(X, O)Y X
Y X Y
Y Y X
X Y F X Y
F=F∩Y Y
(X, d)
(On)n∈N
r∈QOr=R\ {r}
OrRTr∈QOr=R\Q R
(Xn,On)1≤n≤p
X=X1× · · · × XnX
O1× · · · × OnOkXk1≤k≤n
X
R2O1×O2
X∅ F (Fi)i∈I
FJ={j∈I:Fj=X}Ti∈IFi=Tj∈JFj
I J 6=∅Si∈IFi=X J =∅
Si∈IFiX
F=P(X)X
O1O2X\(O1∩O2) = (X\O1)∪(X\O2)
O1O2
X\(O1∩O2)6=X X
O1∩O26=∅τcf
x y
P:R→RF
Rcf F=RP−1(F) = RF P −1(F)
P−1(F)
Rcf FRcf PRcf
g:Rcf →Rcf P
g y ∈Rg−1({y})
Rcf Rg g−1({y})
y∈Rg−1({y})y∈R
FRcf RF=R
R Rcf g

gR R
{y}g g−1
g g−1Rcf
RτRτcf
τR
τRτcf
x∈RO3x(R, τcf )R\O
{n∈N:rn∈R\O}
rn∈O n
(rn)n∈Nx
(E, d)
x∈E U x V
x r > 0B[x, r]⊂V
V⊃B[x, r]∩U B[x, r]∩U U
x E
x E ]0,1[
r∈QV r QV⊃[r−η, r +η]∩Q
η > 0x∈(R\Q)∩]r−η, r +η
(rn)n∈N]r−η, r +η[x
V V
Q
A x ∈A U x E
A U ∩A x A
U A A
x∈A A x U
x E x A U ⊂A
A= [0,1[
R[0,1
2] 0 A x ∈]0,1[
η > 0 [x−η, x +η]⊂A
[x−η, x +η]x A
(x, x0)∈E×E0U U0x x0
U×U0(x, x0)E×E0U×U0
E×E0(x, x0)∈E×E0
E× {x0}E×E0
E× {x0}E
E E0
x∈A U x A η > 0
B[x, η]∩A⊂U y ∈B(x, η)A
(yn)n∈NA y n
yn∈B(x, η)∩A⊂U(yn)n∈NU
U y ∈U⊂A B(x, η)⊂A A
Q