Download Solution of exercise sheet 7

Topology
D-MATH, FS 
Damien Calaque
Solution sheet 
Compa-Open Topology
. Let ϕ : T → C({∗}, T) be defined by ϕ(t) = ∗ 7→ t. Then for any open
set U ⊂ T, ϕ(U) = Ω({∗}, U). ϕ is a bijeion and induces a bijeion on the
topologies, so ϕ is a homeomorphism.
. Let f , g be two continuous funions from S to T, and set d(f , g) =
sups∈S dT (f (s), g(s)). Then because S is compa, d(f , g) is always finite and
so it is a diance on C(S, T).
Let Tco denote the Compa-Open topology on C(S, T) and Td the metric
topology.
Tco ⊂ Td : let K be a compa subset of S, t be an element of T and ε > 0.
Choose a funion f inside Ω(K, BT (t, ε)). Now, because K is compa and
f is continuous, f (K) ⊂ BT (t, ε) is a compa set. This means that we can
find η < ε such that f (K) ⊂ BT (t, η). Then we have
Bd (f , ε − η) ⊂ Ω(K, BT (t, ε)).
Td ⊂ Tco : let f : S → T be a continuous funion, r > 0. Cover T by open
balls BT (ti , r/3) of radius r/3 and pull this open cover back to S via f . Since
S is compa, there is a finite subcover U1 ∪ ... ∪ Un . Then closures Ui are
compa and f (U i ) ⊂ BT (ti , r/2). Therefore f satisfies
\
f ∈
Ω(U i , BT (ti , r/2)) =: N
i∈~1,n
Claim : N ⊂ Bd (f , r). Les g ∈ N and take any point s ∈ S. Since the Ui
cover S, we have s ∈ Ui for some index i. The values f (s) and g(s) satisfy :
f (s) ∈ f (Ui ) ⊂ BT (ti , r/3)
g(s) ∈ g(Ui ) ⊂ g(U i ) ⊂ BT (ti , r/2).
By the triangle inequality, we obtain :
d T (f (s), g(s)) ≤ d T (f (s), ti ) + d T (ti , g(s)) < 5r/6
and since s ∈ S was arbitrary, we conclude d T (f , g) < r.
. Let f , g be two diin continuous funions from S to T. Then there
exis a point s ∈ S such that f (s) , g(s) in T. Because T is Hausdorff, we
can
Topology
D-MATH, FS 
Damien Calaque
Alexandroff Compaification
Because S is locally compa, for any point x ∈ S there exis Kx a compa neighbourhood of x in S. Then in b
S, K and {ω} ∪ Kc are two disjoint
neighborhoods separating x and ω. And as S is already Hausdorff, we deduce that b
S is Hausdorff.
Now let (Ui )i∈I be an open cover of b
S. We can suppose that U1 contains
ω. So there exis a compa subspace K of S such that Kc ∪ {ω} = U1 .
Because K is compa, we can find a finite subset J of I such that (K ∩ Uj )j∈J
S
is a finite cover of K. Then U1 ∪ j∈J Uj is a finite sub-cover of b
S.
Hence it is a compa space.
Proper maps
. Suppose S is compa et let Z be any topological set and F be any closed
set in S × Z. Choose z that is not in the projeion of F on Z. Then for any
s ∈ S there exis an open subsets of S and Z, call them Us , Vs such that
(s, x) ∈ Us × Vs and Us × Vs ∩ F = ∅ because F is closed. Now because S is
compa, S can be covered by only a finite number of these Us , call them
U1 , ..., Un . Then V = ∩1≤i≤n Vi is a neighbourhood of z in Z and the image
of F is a closed subset of Z. Hence the map is closed, meaning that S → {∗}
is proper.
Conversely, if S → {∗} is proper, let (Ui )i∈I be an open cover. We build a
new space b
S = S ∪ {ω} with the following topology : a set F ⊂ b
S is closed if
ω ∈ F or if F ⊂ ∪j∈J Uj where J ⊂ I is finite. Then let Γ = ∆ be the closure of
the diagonal ∆ = {(x, x) ∈ S × b
S |x ∈ S}. By properness, the image of Γ in b
S
is a closed set and as it is not containing ω, it is covered by a finite number
of open sets U1 , ..., Un and so S is compa. Indeed, suppose (x, ω) ∈ Γ for
some x ∈ S. Since (Ui )i∈I is an open cover, x ∈ Ui for some i ∈ I. But Ui
is closed in b
S, which implies that b
S − Ui is an open neighbourhood of ω.
By definition of the produ topology, Ui × (b
S − Ui ) is open in S × b
S. But
b
Ui × ( S − Ui ) ∩ ∆ = ∅, which is a contradiion.
. The properness of the composition of two proper maps follows direly
from the fa that the composition of two closed maps is closed. If f : S → T
and f 0 : S0 → T0 are two proper maps, then f × f 0 is proper because for any
Z, the map f × f 0 × idZ is the composition of two closed maps idS × f 0 × idZ :
S × S0 × Z → S × T0 × Z and f × idT0 × idZ : S × T0 × Z → T × T0 × Z.
. We use a circular proof 1 ⇒ 2 ⇒ 3 ⇒ 1. Suppose f proper, then using
the case Z = {∗}, f is closed. Let t be a point of T, then because f : S → T is
proper, f : f −1 (t) → {t} is also proper. Hence by , f −1 (t) is compa.
Suppose f is closed and that the inverse image of any point is a compa
subset of S. Let K be a compa subspace of T and (Ui )i∈I an open cover of
Topology
D-MATH, FS 
Damien Calaque
f −1 (K). For any t ∈ K, we know that f −1 ({t}) is compa, so there is a finite
sub-cover of it (Ui )i∈It . Let Vt be the union of the Ui for all i ∈ It , because
f is closed, f (Vct )c is an open subset of T. These open sets form a cover of
K, so we can extra a finite family t1 , ..., tn such that (f (Vcti )c )1≤i≤n covers
K. Now (Uj )j∈It for 1 ≤ i ≤ n is a finite sub-cover of S, showing that f −1 (K)
i
is compa.
Finally, suppose that the inverse image of any compa set is compa.
◦
Let (Ki )i∈I be a family of compa neighbourhoods of T such that ∪i∈I ( Ki ) =
T. Then for every i ∈ I, f −1 (Ki ) is a compa set in S ; and they cover S.
Now the composition f −1 (Ki ) → Ki → {∗} is proper, and T is Hausdorff, so
f : f −1 (Ki ) → Ki is also proper. The conclusion is that f itself is proper
because the interior of the f −1 (Ki ) cover S.
We used the following lemma : if f : X → X0 and g : X0 → X00 are such
that g ◦ f is proper and X0 is Hausdorff, then f is proper.
Proof : let ϕ : X → X × X0 be defined by ϕ(x) = (x, f (x)). Then because
X0 is Hausdorff, ϕ(X) is closed in X × X0 . Because g ◦ f is proper and Id X0 is
proper, g ◦ f × Id X0 is proper. Now because ϕ(X) is closed, g ◦ f × Id X0 ◦ ϕ is
proper. But, it happens that this map is also the composition ψ ◦ f where
ψ : X0 → X00 × X0 defined by ψ(y) = (y, g(y)). So we know ψ ◦ f is proper.
But ψ is injeive, so f is proper.
Compa or not compa ?
. This subset of R is not closed, so it cannot be compa.
. A matrix A is orthogonal if and only if t AA = I, so it is closed subset of
2
Rn . Moreover, if A is orthogonal, then ||A|| = 1 so it is bounded. Thus it is
compa.
. Any finite topological space has a finite topology, so it is compa.
. The Cantor set is an interseion of closed sets by conruion, so it is
closed in R. And it lies inside [0, 1], hence it is bounded and thus compa.