8. Tychonoff's theorem and the Banach-Alaoglu theorem - Aarhus ...

8. Tychonoff’s theorem and the Banach-Alaoglu
theorem
Klaus Thomsen matkt@imf.au.dk
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 assumption is that for any collection Uα,α ∈ I,
of sets from S such that X =
α∈I Uα, there is a finite subset
F ⊆ I such that X =
α∈F Uα.
And the conclusion 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
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.
It follows then first that X = Y1 ∪ Y2 ∪ · · · ∪ Yn ∪ ( n
i=1 Vi),
and then that
X = Y1 ∪ Y2 ∪ · · · ∪ Yn ∪ ( n
i=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
α∈I
Xα =
(xα) α∈I : xα ∈ Xα .
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
p −1
α (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.
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
(V) for some open subset V ⊆ Xα.
form p −1
α
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 p −1
β (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
Γ ∩ Sβ = p −1
β (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 by assumption, there is a finite subset F ⊆ J
such that Xβ =
j∈F Uj.
Then p −1
β (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
is weak* compact.
Proof.
K = {l ∈ X ∗ : |l(x)| ≤ 1 ∀x ∈ V }.
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 = {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:
l = sup {|l(x)| : x ∈ X, x ≤ 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 : x < 1} we obtain
Theorem
The unit ball
{l ∈ X ∗ : l ≤ 1}
of X ∗ is compact in the weak ∗ -topology.
Klaus Thomsen 8. Tychonoff’s theorem and the Banach-Alaoglu theorem