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Transcript
M21–651, Introduction to Topology
Professor Walkington
1
Weak Topologies
Definition 1.1 Let X be a set and {Xα }α∈A be a family of topological spaces
and fα : X → Xα . The weak topology on X induces by {fα | α ∈ A} is the
smallest topology on X for which each fα is continuous.
The following lemma summarizes basic properties of the weak topology.
Lemma 1.2 Let X have the weak topology induced by a collection {fα : X →
Xα | α ∈ A}.
1. Sets of the form fα−1 (Uα ) with Uα sub-basic open sets in Xα form a
subbase for the topology on X.
2. If Y is a topological space then f : Y → X is continuous if and only if
fα ◦ f : Y → Xα is continuous.
3. If B ⊂ X, then the subspace topology is induced by the restriction of fα
to B.
4. If each Xα has the weak topology induced by a collection {gαλ : Xα →
Yαλ | λ ∈ Λα } then X has the weak topology given by the collection
{gαλ ◦ fα : X → Yαλ | λ ∈ Λα }.
Proof.
1. A sub-base for the weak topology is
S = {fα−1 (Uα ) | Uα ⊂ Xα is open}.
If Sα is a sub-base for Xα , then any open set Uα in Xα may be written
as
Uα = ∪i∈I ∩nj=1 Sαij
1
with Sαij ∈ Sα , for some index set I and integer n. Then
fα−1 (Uα ) = ∪i∈I ∩nj=1 fα−1 (Sαij ),
which implies
S 0 = {fα−1 (Sα ) | Sα ∈ Sα }
is a also a sub-base for the weak topology.
2. This is a crucial property. If f is continuous, then so too is the composition fα ◦ f : Y → Xα .
Next, suppose each of these functions is continuous. Then a sub-basic
open set in X takes the form S = fα−1 (Uα ) with Uα ⊂ Xα open. Then
f −1 (S) = (f ◦ fα )−1 (Uα ),
being open by hypothesis, shows that the inverse image of a sub-basic
open set is open, so f is continuous.
3. Let Uα ⊂ Xα be open, then
(fα |B )−1 (α) = fα (Uα ) ∩ B.
The set on the left is a sub-basic open set for the weak topology on
B, and the set on the right is a sub-basic open set for the sub-space
topology; hence the two topologies agree.
4. A sub-basic open set for the topology on Xα takes the from Uα =
−1
gαλ
(Vαλ ), where Vαλ ⊂ Yαλ is open. Then a sub-basic open set in X
takes the from
fα−1 (Uα ) = (gαλ ◦ fα )−1 (Vαλ ).
Definition 1.3 Let fα : X → Xα , for each α ∈ A.
1. The evaluation map e : X →
Q
α
Xα is defined by e(x)α = fα (x).
2. The collection {fα | α ∈ A} separates points in X if whenever x 6= y
in X there exists α ∈ A such that fα (x) 6= fα (y).
2
3. The collection {fα | α ∈ A} separates points from closed sets in X
if whenever B is closed in X and x 6∈ B there exists α ∈ A such that
fα (x) 6∈ fα (B).
Recall that the continuous functions from a completely regular space X to
[0, 1] separate points from closed sets, and Urysohn’s theorem shows that all
T1 normal spaces are completely regular.
Definition 1.4 Let X and Y be topological spaces and f : X → Y be injective and continuous and let Z = f (X) have the subspace topology. If
f : X → Z is a homeomorphism then the mapping f : X → Y is a (topological) embedding.
The following theorem is used ubiquitously to characterize various topological
spaces as subspaces of a product of typically simpler spaces (e.g. Xα = [0, 1]
for all α).
Theorem 1.5 Let X and Xα be topological spaces
for each α ∈ A and fα :
Q
X → Xα . Then the evaluation map e : X → α Xα is an embedding if and
only if X has the weak topology induced by the collection {fα | α ∈ A} and
this collection separates points.
Proof. (⇒) This is the easy half. An embedding separates points, thus if
x 6= y it follows that e(x) 6= e(y) so for some α
fα (x) = πα ◦ e(x) 6= πα ◦ e(y) = fα (y).
Next, a sub-basic open subset of e(X) in the subspace topology of the product
takes the from Sα = πα−1 (Uα ) ∩ e(X), with Uα ⊂ Xα open. Since e is an
embedding it follows S = e−1 (Sα ) is a sub-basic open set for X. Then
S = e−1 (Sα ) = e−1 (πα−1 (Uα )) = (πα ◦ e)−1 (Uα ) = fα−1 (Uα ),
shows that S is a sub-basic open set of the weak topology on X. thus
the topology on X is no coarser than the weak topology. Moreover, since
fα = πα ◦ e is a composition of continuous functions each fα is continuous,
so the topology on X contains the weak topology induced by the fα .
3
Q
(⇐) Since the embedding separates points it follows that e : X → α Xα is
injective. Also, since πα ◦ e = fα and the latter is continuous when X has
the weak topology, it follows that πα ◦ e is continuous for all α which implies
e is continuous. It then suffices to show that e is an open map.
Let U = fα−1 (Uα ) be a sub-basic open set in X. Since e is a bijection we have
πα−1 (Uα ) ∩ e(X) =
=
=
=
e ◦ e−1 ◦ πα−1 (Uα )
e ◦ (πα ◦ e)−1 (Uα )
e ◦ fα−1 (Uα )
e(U ).
The set on the left is a sub-basic open set of the subspace e(X).
If X is a topological space the following lemma can be used to determine if
it is the weak topology of a collection {fα | α ∈ A}
Lemma 1.6 A collection {fα : X → Xα | α ∈ A} of continuous functions
on a topological space X separates points from closed sets in X if and only if
the sets fα−1 (V ), for α ∈ A and V open in Xα , form a base for the topology
on X.
Proof. (⇒) Fix U ⊂ X open, and for each x ∈ U let fα (x) be disjoint from
Bx ≡ fα (X \ U ). Then
x ∈ fα−1 (Xα \ Bx ) = X \ fα−1 (Bx ) ⊂ U.
It follows that U = ∪x∈U fα−1 (Xα \ Bx ).
(⇐) Let the sets form a base for the topology, and B ⊂ X be closed and
x 6∈ B. Then x ∈ X \ B is open, so there exists a basic open set such that
x ∈ fα−1 (V ) ⊂ X \ B,
V ⊂ Xα open.
Then f (B) ∈ X \ V and the set on the right is closed, so fα (B) ⊂ X \ V ,
and since fα (x) ∈ V it follows f (x) 6∈ fα (B)
Corollary 1.7 If {fα : X → Xα | α ∈ A} is a collection of continuous
functions on a topological space X which separates points from closed sets,
then the topology on X is the weak topology induced by the fα .
4
Recall that a topological space is T1 if and only if points are closed.
Theorem 1.8 If X is a T1 space and {fα : X → Xα | α ∈ A} is a collection
of continuous functions
Q which separates points from closed sets, then the
evaluation e : X → α Xα is an embedding.
Proof. The corollary shows that X has the weak topology induced by the fα ,
and since X is a T1 space, a collection which separates points from closed
sets also separates of points. Theorem ?? then shows that the evaluation
map is an embedding.
5