Download Math 441 Summer 2009: Infinite Products (§19) Recall from last time

Math 441 Summer 2009: Infinite Products (§19)
Recall from last time:
Given an infinite cartesian product of topological spaces,
Y
X=
Xα ,
α∈Λ
we defined two topologies on X:
• BOX TOPOLOGY: open sets are arbitrary unions of the following basis elements:
(
)
Y
Bbox =
Uα | Uα is open in the topology on Xα for each α ∈ Λ ,
α∈Λ
or, further reduced:
Y
Bbox = {
Bα | Bα is a basis set for the topology on Xα for each α ∈ Λ}
α∈Λ
• PRODUCT TOPOLOGY: open sets are arbitrary unions of the following basis elements:
Y
Bprod = {
Uα | Uα open in Xα for all α ∈ Λ, and Uα = Xα for all but finitely many α ∈ Λ},
α∈Λ
or, further reduced:
Y
Bprod = {
Uα | Uα = Xα for all but finitely many indices α ∈ Λ, and remainingUα basis sets for Xα }
α∈Λ
For example, if X =
Y
R, then:
n≥1
1 1
(− , ) is open in the box topology, but not in the product topology on X, while
n n
n≥1
Y
V = (0, ∞) ×
R is open in both product and box topologies.
U=
Y
n>1
Remarks:
1) Since Bprod ⊂ Bbox , the box topology is finer than the product topology.
(In fact, it is too fine, as we shall see below.)
2) Both the box and the product topologies are extensions of the finite product topology.
Example of why the box topology is ”bad”: Consider the set Rω := R × R × R × ... =
∞
Y
i=1
Note that we can think of this set as all sequences of reals.
Let f : R → Rω be the ”diagonal” map: f (x) = (x, x, x, ...).
Relative to any ”good” topology, such a nice map should be continuous.
But if we endow Rω with the box topology, this map is not continuous!
Proof that f : R → (Rω , Tbox ) is not continuous:
1
R.
Y 1 1
⊂ Rω . Then U is open in the box topology on the co-domain Rω
Proof. Let U :=
− ,
n n
n≥1
(because it is a cartesian product
of open sets in the standard topology on R).
\ 1 1
= {0} is not open in the domain R of the function. This
However, f −1 (U ) =
− ,
n n
n≥1
shows that f is not continuous.
Q: Are the projection maps πj :
Y
Xn → Xj continuous relative to the product topology?
n≥1
How about relative to the box topology?
A: Yes, and yes. In fact, the product topology is precisely the smallest topology on an infinite
product such that all projection maps are continuous (Why? Convince yourself of this!)
Proof: Let
Y j be any index
Y value, and take Uj to be any open set in the space
Y Xj . Then
=
Xn × Uj ×
Xn is open in the product (and box) topologies on
Xn , so the
πj−1 (Uj )
n>j
1≤n<j
n≥1
projection map πj is continuous.
Recall from section 18: f : Z → X × Y is continuous iff both πX ◦ f and πY ◦ f are
continuous.
Is this true for maps into infinite products, under the box topology?
NO. See the diagonal function example above.
It does generalize to infinite products with product topology, however:
THEOREM:
Y
Let (Xn , Tn ) be an indexed family of topological spaces, n ∈ J. Let (
Xn , Tprod ) be
n≥1
the cartesian
product of these spaces, with the product topology. Then a function
Y
f :Z→
Xn is continuous ⇔ πj ◦ f : Z → Xj is continuous, for all j ≥ 1.
n≥1
Proof. The forward direction is easy, and it holds true in both the product and box topologies, since
it relies only on the continuity of the projection maps. Since we proved above that the projection
maps πj are continuous (relative to either topology), if f is continuous then πj ◦ f is a composition
of continuous functions, hence continuous.
The converse
is only true
πj ◦ f are continuous. Consider
Y
Y in the product topology. Suppose allY
f :Z→
Xn . Let U =
Un be a basis set for the codomain
Xn . That is, Un = Xn for all
n≥1
n≥1
n≥1
indices n not in a finite set S = {n1 , n2 , ...nk }. Then the pre-image of U is:
!
!
\
\
\
f −1 (U ) =
(π ◦ f )−1 (Un ) =
(π ◦ f )−1 (Un ) ∩
(π ◦ f )−1 (Un )
n≥1
n∈S
n∈S
/
2
!
=
\
−1
(π ◦ f )
n∈S
(Un )
!
∩
\
n∈S
/
Z
=
\
(π ◦ f )−1 (Un )
n∈S
This is a finite intersection of open sets in the domain Z, so it is open.
Both the box and the product topologies behave well with respect to some
properties:
1) If all of (Xn , Tn ) are Hausdorff, then so is their product (in either topology)
Q
2) If An ⊂ Xn for allQi, then the product (or box) topology on An is the same as the subset
topology induced from Xn with the product (or box) topology.
Y
Y
An =
3)
An , in either topology.
n
n
Proof. In text or left to reader
3