Stonecech
Stonecech
Stonecech
spaces
Jordan Bell
jordan.bell@gmail.com
Department of Mathematics, University of Toronto
June 27, 2014
1
Theorem 3. If X is Hausdorff and A ⊂ X, then A with the subspace topology
is completely
Q regular. If {Xi : i ∈ I} is a family of completely regular spaces,
then i∈I Xi is completely regular.
Proof. Suppose that F is closed in A with the subspace topology and x ∈ A \ F .
There is a closed set G in X with F = G ∩ A. Then x 6∈ G, so there is a
continuous function f : X → [0, 1] satisfying f (x) = 0 and f (F ) = {1}. The
restriction of f to A with the subspace topology is continuous, showing that A
is completely regular. Q
Suppose that F is a closed subset of X = i∈I Xi and that x ∈ X \ F . A
base for the product topology consists of intersections of finitely many sets of
the form πi−1 (Ui ) where i ∈ I and Ui is an open subset of Xi , and because X \ F
is an open neighborhood of x, there is a finite subset J of I and open sets Uj
in Xj for j ∈ J such that
\
x∈ πj−1 (Uj ) ⊂ X \ F.
j∈J
2
2 Initial topologies
Suppose that X is a set, Xi , i ∈ I, are topological spaces, and fi : X → Xi are
functions. The initial topology on X induced by {fi : i ∈ I} is the coarsest
topology on X such that each fi is continuous. A subbase for the initial topology
is the collection of those sets of the form fi−1 (Ui ), i ∈ I and Ui open in Xi .
If fi :QX → Xi , i ∈ I, are functions, the evaluation map is the function
e : X → i∈I Xi defined by
3
A subbase for the topology of X consists of those sets of the form V =
fi−1 (Ui ), i ∈ I and Ui open in Xi . As fi = pi ◦ e we can write this as
4
3 Bounded continuous functions
For any set X, we denote by `∞ (X) the set of all bounded functions X → R, and
we take as known that `∞ (X) is a Banach space with the supremum norm
which is continuous and ≥ 0, and satisfies f −1 (af , ∞) = gf−1 (0, ∞), so that
\
x∈ gf−1 (0, ∞) ⊂ U.
f ∈J
5
Q
Define g = f ∈J gf , whichQis continuous because each factor is continuous. This
function satisfies g(x) = f ∈J gf (x) > 0 because this is a product of finitely
many factors each of which are > 0. If y ∈ g −1 (0, ∞) then y ∈ f ∈J gf−1 (0, ∞) ⊂
T
4 Compactifications
In §1 we talked about the one-point compactification of a locally compact Haus-
dorff space. A compactification of a topological space X is a pair (K, h) where
(i) K is a compact Hausdorff space, (ii) h : X → K is an embedding, and (iii)
h(X) is a dense subset of K. For example, if X is a compact Hausdorff space
then (X, idX ) is a compactification of X, and if X is a locally compact Hausdorff
space, then the one-point compactification X ∗ = X ∪ {∞}, where ∞ is some
symbol that does not belong to X, together with the inclusion map X → X ∗ is
a compactification.
Suppose that X is a topological space and that (K, h) is a compactification
of X. Because K is a compact Hausdorff space it is normal, and then Urysohn’s
lemma tells us that K is completely regular. But K is Hausdorff, so in fact
K is Tychonoff. A subspace of a Tychonoff space is Tychonoff, so h(X) with
the subspace topology is Tychonoff. But X and h(X) are homeomorphic, so
X is Tychonoff. Thus, if a topological space has a compactification then it is
Tychonoff.
6
In Theorem 8 we proved that any Tychonoff space can be embedded into a
cube. Here review our proof of this result. Let X be a Tychonoff space, and
for each f ∈ Cb (X) let If = [− kf k∞ , kf k∞ ], so that f : X → If is continuous,
and the family of theseQfunctions separates points in X. The evaluation map for
this family is e : X → f ∈Cb (X) If defined by (πf ◦ e)(x) = f (x) for f ∈ Cb (X),
Q
and Theorem 6 tells us that e : X → f ∈Cb (X) If is an embedding. Because
each interval If is a compact Hausdorff space (we remarkQ that if f = 0 then
If = {0}, which is indeed compact), the product f ∈Cb (X) If is a compact
Hausdorff space, and henceQ any closed subset of it is compact. We define βX to
be the closure of e(X) in f ∈Cb (X) If , and the Stone-Čech compactification
of X is the pair (βX, e), and what we have said shows that indeed this is a
compactification of X.
The Stone-Čech compactification of a Tychonoff space is useful beyond dis-
playing that every Tychonoff space has a compactification. We prove in the
following that any continuous function from a Tychonoff space to a compact
Hausdorff space factors through its Stone-Čech compactification.4
Theorem 9. If X is a Tychonoff space, K is a compact Hausdorff space, and
φ : X → K is continuous, then there is a unique continuous function Φ : βX →
K such that φ = Φ ◦ e.
Proof. K is Tychonoff because aQcompact Hausdorff space is Tychonoff, so
the evaluation map eK : K → g∈Cb (K) Ig is an embedding. Write F =
Q Q
f ∈Cb (X) If , G = g∈Cb (K) Ig , and let pf : F → If , qg : G → Ig be the
projection maps.
We define H : F → G for t ∈ F by (qg ◦ H)(t) = t(g ◦ φ) = pg◦φ (t). For each
g ∈ G, the map qg ◦ H : F → Ig◦φ is continuous, so H is continuous.
For x ∈ X, we have
so
H ◦ e = eK ◦ φ. (1)
On the one hand, because K is compact and eK is continuous, eK (K) is
compact and hence is a closed subset of G (G is Hausdorff so a compact subset
is closed). From (1) we know H(e(X)) ⊂ eK (K), and thus
7
On the other hand, because βX is compact and H is continuous, H(βX) is
compact and hence is a closed subset of G. As e(X) is dense in βX and H is
continuous, H(e(X)) is dense in H(βX), and thus
H(e(X)) = H(βX) = H(βX).
Therefore we have
H(βX) ⊂ eK (K).
Let h be the restriction of H to βX, and define Φ : βX → K by Φ = e−1
K ◦ h,
which makes sense because eK : K → eK (K) is a homeomorphism and h takes
values in eK (K). Φ is continuous, and for x ∈ X we have, using (1),
−1
(Φ ◦ e)(x) = (eK ◦ h ◦ e)(x) = (e−1
K ◦ H ◦ e)(x) = φ(x),
showing that Φ ◦ e = φ.
If Ψ : βX → K is a continuous function satisfying f = Ψ ◦ e, let y ∈ e(X).
There is some x ∈ X such that y = e(x), and f (x) = (Ψ ◦ e)(x) = Ψ(y),
f (x) = (Φ ◦ e)(x) = Φ(y), showing that for all y ∈ e(X), Ψ(y) = Φ(y). Since Ψ
and Φ are continuous and are equal on e(X), which is a dense subset of βX, we
get Ψ = Φ, which completes the proof.
If X is a Tychonoff space with Stone-Čech compactification (βX, e), then
because βX is a compact space, C(βX) with the supremum norm is a Banach
space. We show in the following that the extension in Theorem 9 produces an
isometric isomorphism Cb (X) → C(βX).
Theorem 10. If X is a Tychonoff space with Stone-Čech compactification
(βX, e), then there is an isomorphism of Banach spaces Cb (X) → C(βX).
Proof. Let f, g ∈ Cb (X), let α be a scalar, and let K = [−|α| kf k−kgk , |α| kf k+
kgk], which is a compact set. Define φ = αf + g, and then Theorem 9 tells us
that there is a unique continuous function F : βX → K such that f = F ◦ e,
a unique continuous function G : βX → K such that g = G ◦ e, and a unique
continuous function Φ : βX → K such that φ = Φ ◦ e. For y ∈ e(X) and x ∈ X
such that y = e(x),
Φ(y) = φ(x) = αf (x) + g(x) = αF (y) + G(y).
Since Φ and αF + G are continuous functions βX → K that are equal on the
dense set e(X), we get Φ = αF + G. Therefore, the map that sends f ∈ Cb (X)
to the unique F ∈ C(βX) such that f = F ◦ e is linear.
Let f ∈ Cb (X) and let F be the unique element of C(βX) such that f = F ◦e.
For any x ∈ X, |f (x)| = |(F ◦ e)(x)|, so
kf k∞ = sup |f (x)| = sup |(F ◦ e)(x)| = sup |F (y)|.
x∈X x∈X y∈e(X)
8
so kf k∞ = kF k∞ , showing that f 7→ F is an isometry.
For Φ ∈ C(βX), define φ = Φ ◦ e. Φ is bounded so φ is also, and φ is a
composition of continuous functions, hence φ ∈ Cb (X). Thus φ 7→ Φ is onto,
completing the proof.
9
Suppose that fn is a sequence of elements of C0 (X) that converges to some
f ∈ Cb (X). For > 0, there is some n such that n ≥ n implies that
kfn − f k∞ < 2 , that is,
sup |fn (x) − f (x)| < .
x∈X 2
Cc (X) ⊂ C0 (X),
Proposition 4.35.
6 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, p. 131,
Lemma 4.32.
10
fn = gn f ∈ Cc (X). (A product of continuous functions is continuous, and
because f is bounded and gn has compact support, gn f has compact support.)
For x ∈ Cn , fn (x) − f (x) = (gn (x) − 1)f (x) = 0, and for x ∈ X \ Cn , |fn (x) −
f (x)| = |gn (x) − 1||f (x)| ≤ 1 · n1 . Therefore
1
kfn − f k∞ ≤ ,
n
and hence fn is a sequence in Cc (X) that converges to f , showing that Cc (X)
is dense in C0 (X).
If X is a Hausdorff space, then we prove that Cc (X) is a linear subspace of
C0 (X). When X is a locally compact Hausdorff space then combined with the
above this shows that Cc (X) is a dense linear subspace of C0 (X).
Lemma 13. Suppose that X is a Hausdorff space. Then Cc (X) is a linear
subspace of C0 (X).
Proof. If f, g ∈ Cc (X) and α is a scalar, let K = supp f ∪ supp g, which is
a union of two compact sets hence compact. If x ∈ X \ K, then f (x) = 0
because x 6∈ supp f and g(x) = 0 because x 6∈ supp g, so (αf + g)(x) = 0.
Therefore {x ∈ X : (αf + g)(x) 6= 0} ⊂ K and hence supp (αf + g) ⊂ K. But
as X is Hausdorff, K being compact implies that K is closed in X, so we get
supp (αf + g) ⊂ K. Because supp (αf + g) is closed and is contained in the
compact set K, it is itself compact, so αf + g ∈ Cc (X).
Let X be a topological space, and for x ∈ X define δx : Cb (X) → R by
δx (f ) = f (x). For each x ∈ X, δx is linear and |δx (f )| = |f (x)| ≤ kf k∞ , so
δx is continuous and hence belongs to the dual space Cb (X)∗ . Moreover, the
constant function f (x) = 1 shows that kδx k = 1. We define ∆ : X → Cb (X)∗ by
∆(x) = δx . Suppose that xi is a net in X that converges to some x ∈ X. Then
for every f ∈ Cb (X) we have f (xi ) → f (x), and this means that δxi weak-*
converges to δx in Cb (X)∗ . This shows that with Cb (X)∗ assigned the weak-*
topology, ∆ : X → Cb (X)∗ is continuous. We now characterize when ∆ is an
embedding.7
Theorem 14. Suppose that X is a topological space and assign Cb (X)∗ the
weak-* topology. Then the map ∆ : X → ∆(X) is a homeomorphism if and
only if X is Tychonoff, where ∆(X) has the subspace topology inherited from
Cb (X)∗ .
Proof. Suppose that X is Tychonoff. If x, y ∈ X are distinct, then there is some
f ∈ Cb (X) such that f (x) = 0 and f (y) = 1, and then δx (f ) = 0 6= 1 = δy (f ),
so ∆(x) 6= ∆(y), showing that ∆ is one-to-one. To show that ∆ : X → ∆(X)
is a homeomorphism, it suffices to prove that ∆ is an open map, so let U be an
open subset of X. For x0 ∈ U , because X \ U is closed there is some f ∈ Cb (X)
such that f (x0 ) = 0 and f (X \ U ) = {1}. Let
V1 = {µ ∈ Cb (X)∗ : µ(f ) < 1}.
7 John B. Conway, A Course in Functional Analysis, second ed., p. 137, Proposition 6.1.
11
This is an open subset of Cb (X)∗ as it is the inverse image of (−∞, 1) under the
map µ 7→ µ(f ), which is continuous Cb (X)∗ → R by definition of the weak-*
topology. Then
V = V1 ∩ ∆(X) = {δx : f (x) < 1}
is an open subset of the subspace ∆(X), and we have both δx0 ∈ V and V ⊂
∆(U ). This shows that for any element δx0 of ∆(U ), there is some open set V in
the subspace ∆(X) such that δx0 ∈ V ⊂ ∆(U ), which tells us that ∆(U ) is an
open set in the subspace ∆(U ), showing that ∆ is an open map and therefore a
homeomorphism.
Suppose that ∆ : X → ∆(X) is a homeomorphism. By the Banach-Alaoglu
theorem we know that the closed unit ball B1 in Cb (X)∗ is compact. (We remind
ourselves that we have assigned Cb (X)∗ the weak-* topology.) That is, with the
subspace topology inherited from Cb (X)∗ , B1 is a compact space. It is Hausdorff
because Cb (X)∗ is Hausdorff, and a compact Hausdorff space is Tychonoff. But
∆(X) is contained in the surface of B1 , in particular ∆(X) is contained in B1 and
hence is itself Tychonoff with the subspace topology inherited from B1 , which is
equal to the subspace topology inherited from Cb (X)∗ . Since ∆ : X → ∆(X) is a
homeomorphism, we get that X is a Tychonoff space, completing the proof.
The following result shows when the Banach space Cb (X) is separable.8
Theorem 15. Suppose that X is a Tychonoff space. Then the Banach space
Cb (X) is separable if and only if X is compact and metrizable.
Proof. Assume that X is compact and metrizable, with a compatible metric d.
For each n ∈ N there are open balls Un,1 , . . . , Un,Nn of radius n1 that cover X.
As X is metrizable it is normal, so there is a partition of unity subordinate
to the cover {Un,k : 1 ≤ k ≤ Nn }.9 That is, there are continuous functions
PNn
fn,1 , . . . , fn,Nn : X → [0, 1] such that k=1 fn,k = 1 and such that x ∈ X \ Un,k
implies that fn,k (x) = 0. Then {fn,k : n ∈ N, 1 ≤ k ≤ Nn } is countable,
so its span D over Q is also countable. We shall prove that D is dense in
C(X) = Cb (X), which will show that Cb (X) is separable.
Let f ∈ C(X) and let > 0. Because (X, d) is a compact metric space, f is
uniformly continuous, so there is some δ > 0 such that d(x, y) < δ implies that
|f (x) − f (y)| < 2 . Let n ∈ N be > 2δ , and for each 1 ≤ k ≤ Nn let xk ∈ Un,k .
For each k there is some αk ∈ Q such that |αk − f (xk )| < 2 , and we define
Nn
X
g= αk fn,k ∈ D.
k=1
PNn PNn
Because fn,k = 1 we have f = k=1
k=1 f fn,k . Let x ∈ X, and then
N N
Xn X n
12
For each 1 ≤ k ≤ Nn , either x ∈ Un,k or x 6∈ Un,k . In the first case, since x and
xk are then in the same open ball of radius n1 , d(x, xk ) < n2 < δ, so
|f (x) − αk | ≤ |f (x) − f (xk )| + |f (xk ) − αk | < + = .
2 2
In the second case, fn,k (x) = 0. Therefore,
Nn
X Nn
X
|f (x) − αk |fn,k (x) ≤ fn,k (x) = ,
k=1 k=1
showing that |f (x) − g(x)| ≤ . This shows that D is dense in C(X), and
therefore that Cb (X) = C(X) is separable.
Suppose that Cb (X) is separable. Because X is Tychonoff, by Theorem
10 there is an isometric isomorphism between the Banach spaces Cb (X) and
C(βX), where (βX, e) is the Stone-Čech compactification of X. Hence C(βX)
is separable. But it is a fact that a compact Hausdorff space Y is metrizable
if and only if the Banach space C(Y ) is separable.10 (This is proved using the
Stone-Weierstrass theorem.) As βX is a compact Hausdorff space and C(βX)
is separable, we thus get that βX is metrizable.
It is a fact that if Y is a Banach space and B1 is the closed unit ball in the
dual space Y ∗ , then B1 with the subspace topology inherited from Y ∗ with the
weak-* topology is metrizable if and only if Y is separable.11 Thus, the closed
unit ball B1 in Cb (X)∗ is metrizable. Theorem 14 tells us there is an embedding
∆ : X → B1 , and B1 being metrizable implies that ∆(X) is metrizable. As
∆ : X → ∆(X) is a homeomorphism, we get that X is metrizable.
Because βX is compact and metrizable, to prove that X is compact and
metrizable it suffices to prove that βX\e(X) = ∅, so we suppose by contradiction
that there is some τ ∈ βX \ e(X). e(X) is dense in βX, so there is a sequence
xn ∈ X, for which we take xn 6= xm when n 6= m, such that e(xn ) → τ . If xn
had a subsequence xa(n) that converged to some y ∈ X, then e(xa(n) ) → e(y)
and hence e(y) = τ , a contradiction. Therefore the sequence xn has no limit
points, so the sets A = {xn : n odd} and B = {xn : n even} are closed and
disjoint. Because X is metrizable it is normal, hence by Urysohn’s lemma there
is a continuous function φ : X → [0, 1] such that φ(a) = 0 for all a ∈ A and
φ(b) = 1 for all b ∈ B. Then, by Theorem 9 there is a unique continuous
Φ : X → [0, 1] such that φ = Φ ◦ e. Then we have, because a subsequence of a
10 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-
13
convergent sequence has the same limit,
Φ(τ ) = Φ lim e(xn )
n→∞
= Φ lim e(x2n+1 )
n→∞
= lim (Φ ◦ e)(x2n+1 )
n→∞
= lim φ(x2n+1 )
n→∞
= 0,
and likewise
Φ(τ ) = lim φ(x2n ) = 1,
n→∞
14
be an algebra isomorphism f : A → B such that f (a∗ ) = f (a)∗ for all a ∈ A.
−1
It follows that kf k ≤ 1 and because
−1
f is bijective, the inverse f is a C ∗ -
algebra homomorphism, giving
f
≤ 1 and therefore kf k = 1. Thus, an
isomorphism of C ∗ -algebras is an isometric isomorphism.
Suppose that A is a commutative C ∗ -algebra, which we do not assume to
be unital. A character of A is a nonzero algebra homomorphism A → C.
We denote the set of characters of A by σ(A), which we call the Gelfand
spectrum of A. We make some assertions in the following text that are proved
in Folland.13 It is a fact that for every h ∈ σ(A), khk ≤ 1, so σ(A) is contained
in the closed unit ball of A∗ , where A∗ denotes the dual of the Banach space A.
Furthermore, one can prove that σ(A) ∪ {0} is a weak-* closed set in A∗ , and
hence is weak-* compact because it is contained in the closed unit ball which we
know to be weak-* compact by the Banach-Alaoglu theorem. We assign σ(A)
the subspace topology inherited from A∗ with the weak-* topology. Depending
on whether 0 is or is not an isolated point in σ(A) ∪ {0}, σ(A) is a compact
or a locally compact Hausdorff space; in any case σ(A) is a locally compact
Hausdorff space.
The Gelfand transform is the map Γ : A → C0 (σ(A)) defined by Γ(a)(h) =
h(a); that Γ(a) is continuous follows from σ(A) having the weak-* topology, and
one proves that in fact Γ(a) ∈ C0 (σ(A)).14 The Gelfand-Naimark theorem15
states that Γ : A → C0 (σ(A)) is an isomorphism of C ∗ -algebras.
It can be proved that two commutative C ∗ -algebras are isomorphic as C ∗ -
algebras if and only if their Gelfand spectra are homeomorphic.16
7 Multiplier algebras
An ideal of a C ∗ -algebra A is a closed linear subspace I of A such that IA ⊂ I
and AI ⊂ I. An ideal I is said to be essential if I ∩ J 6= {0} for every nonzero
ideal J of A. In particular, A is itself an essential ideal.
Suppose that A is a C ∗ -algebra. The multiplier algebra of A, denoted
M (A), is a C ∗ -algebra containing A as an essential ideal such that if B is a C ∗ -
algebra containing A as an essential ideal then there is a unique homomorphism
of C ∗ -algebras π : B → M (A) whose restriction to A is the identity. We have
not shown that there is a multiplier algebra of A, but we shall now prove that
this definition is a universal property: that any C ∗ -algebra satisfying the
definition is isomorphic as a C ∗ -algebra to M (A), which allows us to talk about
“the” multiplier algebra rather than “a” multiplier algebra.
Suppose that C is a C ∗ -algebra containing A as an essential ideal such that
if B is a C ∗ -algebra containing A as an essential ideal then there is a unique
C ∗ -algebra homomorphism π : B → C whose restriction to A is the identity.
13 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 12, §1.3.
14 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 15.
15 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 16, Theorem 1.31.
16 José M. Gracia-Bondı́a, Joseph C. Várilly and Héctor Figueroa, Elements of Noncommu-
15
Hence there is a unique homomorphism of C ∗ -algebras π1 : C → M (A) whose
restriction to A is the identity, and there is a unique homomorphism of C ∗ -
algebras π2 : M (A) → C whose restriction to A is the identity. Then π2 ◦ π1 :
C → C and π1 ◦ π2 : M (A) → M (A) are homomorphisms of C ∗ -algebras
whose restrictions to A are the identity. But the identity maps idC : C → C
and idM (A) : M (A) → M (A) are also homomorphisms of C ∗ -algebras whose
restrictions to A are the identity. Therefore, by uniqueness we get that π2 ◦π1 =
idC and π1 ◦ π2 = idM (A) . Therefore π1 : C → M (A) is an isomorphism of C ∗ -
algebras.
One can prove that if A is unital then M (A) = A.17 It can be proved that
for any C ∗ -algebra A, the multiplier algebra M (A) is unital.18 For a locally
compact Hausdorff space X, it can be proved that M (C0 (X)) = Cb (X).19 This
last assertion is the reason for my interest in multiplier algebras. We have
seen that if X is a locally compact Hausdorff space then Cc (X) is a dense
linear subspace of C0 (X), and for any topological space C0 (X) is a closed linear
subspace of Cb (X), but before talking about multiplier algebras we did not have
a tight fit between the C ∗ -algebras C0 (X) and Cb (X).
16