Download 8. Tychonoff`s theorem and the Banach-Alaoglu theorem

8. Tychonoff’s theorem and the Banach-Alaoglu
theorem
Klaus Thomsen
[email protected]
Institut for Matematiske Fag
Det Naturvidenskabelige Fakultet
Aarhus Universitet
September 2005
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
We read in W. Rudin: Functional Analysis
Covering Chapter 3, Appendix A2 and A3 and Theorem 3.15.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
Alexander’s subbase theorem
Recall that a subbase of a topology is a collection S of open sets
such that every open set is the union of sets that are finite
intersections of elements from S.
Theorem
(Alexander’s subbase theorem) If S is a subbase for the topology of
a space X , and any cover of X by elements of S has a finite
subcover, then X is compact.
In more detail: The assumptionSis that for any collection Uα , α ∈ I ,
of sets from S such that
S X = α∈I Uα , there is a finite subset
F ⊆ I such that X = α∈F Uα .
And the conclusion S
is that for any collection Wα , α ∈ I , of open
sets such that
X
=
α∈I Wα , there is a finite subset F ⊆ I such
S
that X = α∈F Wα .
In short: ’It suffice to check the finite subcover condition with
elements from a subbase’.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The proof of Alexander’s subbase theorem
Proof.
The proof is by contradiction; that is, we assume that X is not
compact, and deduce that there is then an S-cover of X without
any finite subcover.
Let P be the collection of all open covers of X without any finite
subcover; P is non-empty by assumption.
We partial order P by inclusion, and Ω be a maximal totally
ordered subset (which exists by Hausdorff’s maximality theorem).
Let Γ be the union of all members of Ω.
Then
a) Γ is an open cover of X ,
b) Γ has no finite subcover, and
c) Γ ∪ {V } has a finite subcover for every open V ∈
/ Γ.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The proof of Alexander’s subbase theorem
Proof.
a) is obvious.
b) follows because any finite collection from Γ is contained in some
element of Ω; this follows because Ω is totally ordered.
c) follows from the maximality of Ω; if Γ ∪ {V } does not have a
finite subcover for some open V ∈
/ Γ, we could add {V } ∪ Γ to Ω
to get a totally ordered subset of P which is strictly larger than Ω.
We finish the proof by using a)-c) to show that Γ ∩ S is a cover of
X without any finite subcover. (!)
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The proof of Alexander’s subbase theorem
It follows immediately from b) that Γ ∩ S has no finite subcover. It
remains now only to show that Γ ∩ S covers X .
Assume that it does not, i.e. that there is an x ∈ X which is not
contained in any member of Γ ∩ S. Then x ∈ W for some W ∈ Γ.
Since S is a subbase there is a finite collection V1 , V2 , . . . , Vn ∈ S
such that
x ∈ V1 ∩ V2 ∩ · · · ∩ Vn ⊆ W .
Since x is not covered by Γ ∩ S, Vi ∈
/ Γ for each i.
It follows from c) that for each i, there is a union Yi of finitely
many elements of Γ such that X = Vi ∪ Yi .
T
It follows then first that X = Y1 ∪ Y2 ∪ · · · ∪ Yn ∪ ( ni=1 Vi ),
and then that
T
X = Y1 ∪ Y2 ∪ · · · ∪ Yn ∪ ( ni=1 Vi ) ⊆ Y1 ∪ Y2 ∪ · · · ∪ Yn ∪ W ,
which contradicts b).
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
Tychonoff’s theorem
Recall that the cartesian product of a collection Xα , α ∈ I , of
topological spaces is the set
Y
Xα = (xα )α∈I : xα ∈ Xα .
α∈I
Q
Let pβ : α∈I
Xα → Xβ be the canonical surjection
pβ (xα )α∈I = xβ .
The cartesian product is equipped with the product topology which
by definition is the topology for which the sets
pα−1 (V ), α ∈ I , V ⊆ Xα open
is a subbase which we denote by S in the following.
Theorem
If X is the cartesian of any nonempty collection of compact spaces,
then X is compact.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The proof of Tychonoff’s theorem
Proof.
Q
To prove that X = α∈I Xα is compact when each Xα is we must
show that any open cover Γ of X has a finite subcover.
By Alexander’s subbase theorem we may assume that the sets in Γ
come from S - the canonical subbase for the product topology.
For each α ∈ I we let Sα denote the elements of S which are of the
form pα−1 (V ) for some open subset V ⊆ Xα .
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The proof of Tychonoff’s theorem
Proof.
Note first that there must be a β ∈ I such that Γ ∩ Sβ covers X ;
indeed, if not we have a point yβ ∈ X for each β such that yβ is
not contained in any element of Γ ∩ Sβ . Set xβ = pβ (yβ ) ∈ Xβ .
Since each element of Γ ∩ Sβ has the form pβ−1 (W ) for some open
W ⊆ Xβ , we see that xβ is not contained in pβ (A) for any
A ∈ Γ ∩ Sβ .
But then x = (xβ )β∈I is an element of X which is not contained in
any element of Γ; contradicting that Γ is a cover by assumption.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The proof of Tychonoff’s theorem
Proof.
Let β ∈ I be an index such that Γ ∩ Sβ covers X .
Write
n
o
Γ ∩ Sβ = pβ−1 (Uj ) : j ∈ J ,
for some collection {Uj : j ∈ J} of open sets in Xβ .
Note that {Uj : j ∈ J} must cover Xβ because pβ (X ) = Xβ , and
Γ ∩ Sβ covers X .
Since Xβ is compact
S by assumption, there is a finite subset F ⊆ J
such that Xβ = j∈F Uj .
o
n
Then pβ−1 (Uj ) : j ∈ F is a finite subcover of Γ ∩ Sβ !!!
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The Banach-Alaoglu theorem
Thyconoff’s theorem will now be used to prove the following:
Theorem
(Theorem 3.15) Let X be a topological vector space and V an
open neighborhood of X . Then
K = {l ∈ X ∗ : |l (x)| ≤ 1 ∀x ∈ V } .
is weak* compact.
Proof.
By continuity of scalar multiplication there is for each x ∈ X a
positive number γ(x) > 0 such that x ∈ γ(x)V .
Hence |l (x)| ≤ γ(x) for all l ∈ K .
Let Dx =Q{z ∈ Φ : |z| ≤ γ(x)} which is a compact subset of Φ.
Let P = x∈X Dx be the cartesian product, and observe that P is
compact by Thyconoff’s theorem.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The Banach-Alaoglu theorem
Proof.
The elements of P are the functions f : X → Φ with the property
that |f (x)| ≤ γ(x) for all x ∈ X . Thus K ⊆ X ∗ ∩ P.
The proof is completed by showing (i) that the weak* topology on
K is the same as the relative topology which K inherites from the
product topology of P and (ii) K is closed in P.
Both items are straightforward, but tedious to prove. The details
are in Rudins book. We omit them here.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The Banach-Alaoglu theorem
Completing a normed space - an exercise!
When X is a normed vector space - say over C - the dual X ∗ of X
is also a normed space in the dual norm:
kl k = sup {|l (x)| : x ∈ X , kxk ≤ 1} .
First evaluation exercise:
(a) Show that X ∗ is a Banach space.
(b) Let X ∗∗ be the bidual of X , i.e. X ∗∗ = (X ∗ )∗ . Show that the
map Λ : X → X ∗∗ given by
Λ(x)(l ) = l (x),
is a linear isometry.
(c) Prove that Λ(X ) is dense in X ∗∗ for the X ∗ -topology of X ∗∗ .
(d) Interpretate and prove the following statement:
For every normed vector space X there is a Banach space X which
contains an isometric copy of X as a dense subspace. X is unique
up to isometric isomorphisms.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem
The Banach-Alaoglu theorem
If we apply Theorem 3.15 with V = {x ∈ X : kxk < 1} we obtain
Theorem
The unit ball
{l ∈ X ∗ : kl k ≤ 1}
of X ∗ is compact in the weak∗ -topology.
Klaus Thomsen
8. Tychonoff’s theorem and the Banach-Alaoglu theorem