Download Compactness Equivalence and Application to Proper Maps

Equivalent Characterization of Compactness with Application to
Proper Maps
Matt Rosenzweig
Abstract
We define a topological space K to be compact if and only if every open cover of K contains a finite
subcover. We prove a theorem which gives an equivalent criterion for a space to be compact in terms of the
closedness of the projection map πZ : K × Z → Z. We then use this theorem to prove that a proper map is
closed and has compact fibers.
Most references on topological spaces seem to define a compact space in terms of open coverings of a space
having finite subcoverings. Formally, we say that
S a collection of open sets {Ui }i∈I , indexed by some set I, is an
open covering of a topological space K if K ⊂ i∈I Ui . We define K to be compact if there exist finitely many
Sn
open sets Ui1 , · · · , Uin in the open covering such that K ⊂ j=1 Uij .
There exist a number of equivalent definitions of a compact space, which we briefly mention here. A topological
space K is compact if and only if K has the finite intersection
T property: if {Ei }i∈I is a collection of a closed sets
indexed by a set I such that, for any finite subset I0 ⊂ I, i∈I0 Ei 6= ∅, then
\
Ei 6= ∅
i∈I
Another equivalent definition is in terms of nets, also known as Moore-Smith sequences. A topological space K
is compact if and only if every net (xα )A in K has a convergent subnet.
We now prove
Theorem 1. A topological space K is compact if and only, for every topological space Z, the projection map
πZ : K × Z → Z is closed, where K × Z is endowed with the product topology.
Proof. We first prove the ⇒ direction. Suppose K is compact and Z is some topological space. If Z = ∅, then
X × Z = ∅, and there is nothing to prove. So assume that Z 6= ∅. Let E ⊂ K × Z be a closed set, such
that πZ (E) ( Z. Choose a ∈ Z \ πZ (E), so that K × {a} ⊂ E c . Since E c is closed, for all (k, a) ∈ K × {a},
there exists an open neighborhood in E c containing (k, a), which we can take to be a basic open neighborhood
(Vk × Ua,k ) ⊂ E c . The collection {Vk }k∈K is an open covering of K, so there exists a finite subcover Vk1 , · · · , Vkn .
For each k ∈ K, a ∈ Ua,k and Ua,k ∩ πZ (E) = ∅, which implies that
n
\
Ua :=
Ua,kj ⊂ Z
j=1
is an open neighborhood of a. Hence,
K × Ua =
n
[
j=1
Vkj × Ua ⊂
n
[
Vkj × Ua,kj ⊂ E c =⇒ Ua ⊂ πZ (E)c
j=1
Since a ∈ Z \ πZ (E) was arbitrary, we have Z \ πZ (E) is open, hence πZ (E) is closed.
We now prove the ⇐ direction. Let {Ui }i∈I , where I is an indexing set, be an open cover of K. Set
K 0 := K ∪ {∞}, S
where ∞ ∈
/ K. We define a topology τK 0 on K 0 as follows: E ⊂ Y 0 is closed if and only if
{∞} ⊂ E or E ⊂ i∈I0 Ui where I0 is finite. We verify that τK 0 indeed defines a topology.
1. K 0 3 ∞ ⇒ K 0 ∈ τK 0 ; ∅ ⊂ Ui ∀i ∈ I ⇒ ∅ ∈ τK 0 .
2. S
Let {Eα }α be a collection of closed sets. TConsider
i∈I0 Ui . Otherwise, ∞ ∈ Eα ∀α ⇒ ∞ ∈
α Eα .
T
α
Eα . If there exists α, Eα ⊂
S
i∈I0
Ui ⇒
3. S
Let E1 , E2 be closed sets. If either E1 3 ∞ or E2 3 ∞,
S then E1 ∪ E2 3 ∞. Otherwise, E1 ⊂
U
,
for
finite
sets
I
and
I
,
so
that
E
∪
E
⊂
0
1
1
2
i∈I1 i
i∈I0 ∪I1 Ui .
1
S
T
i∈I0
α
Eα ⊂
Ui , E2 ⊂
Consider the closed set {(x, x) ⊂ K × K 0 : x ∈ K}. Suppose (x, ∞) ∈ {(x, x) ⊂ K × K 0 : x ∈ K} for some
x ∈ K. Since {Ui }i is an open cover, x ∈ Ui for some i ∈ I. Ui is closed in K 0 , which implies that K 0 \ Ui 3 ∞
is open. By definition of the product topology, Ui × (K 0 \ Ui ) is open in K × K 0 . But Ui × (K 0 \ Ui ) ∩
{(x, x) ⊂ K × K 0 : x ∈ K} = ∅, which is a contradiction. So
∞∈
/ {(x, x) ⊂ K × K 0 : x ∈ K} =⇒ πK 0 ({(x, x) ⊂ K × K 0 : x ∈ K}) = K,
which is closed in K 0 , since πK 0 is a closed mapping by hypothesis. We conclude that K ⊂
finite subset I0 ⊂ I, and therefore K is compact.
S
i∈I0
Ui , for some
The preceding theorem proves to be a useful technical, as well as interesting, mathematically result for giving
an equivalent characterization of proper maps of topological spaces in terms of the fiber product of spaces. We
briefly review these notions now.
Suppose that X, Y , and Z are topological spaces, and that we have continuous maps f : X → Z and
g : Y → Z. We define the fiber product X ×Z Y to be the set
X ×Z Y := {(x, y) ∈ X × Y : f (x) = g(y) ∈ Z} ⊂ X × Y
We regard X ×Z Y as a topological space by equipping it with the subspace topology induced by the product
topology of X × Y .
We say that a continuous map f : X → Y is proper if, for every continuous map g : Y 0 → Y , the projection
map X ×Y Y 0 → Y 0 is closed.
Theorem 2. A continuous map f : X → Y is proper if and only if f is closed and the fiber f −1 ({y}) is compact,
for each y ∈ Y .
Proof. We first prove the ⇒ direction. To see that f is closed, fix a closed subset K ⊂ X. Let id : Y → Y be
the identity mapping and consider the fiber product X ×Y Y with respect to f and id. K × Y is a closed subset
of X × Y , so by definition of the topology on X ×Y Y , Then
(K × Y ) ∩ (X ×Y Y ) = {(x, y) ∈ X × Y : x ∈ K, f (x) = y} =: E
is a closed subset of X ×Y Y . Let π : X ×Y Y → Y denote the projection map onto Y . By hypothesis that f is
proper, π(E) ⊂ Y is closed. To see that π(E) = f (K), observe that
y ∈ π(E) ⇐⇒ ∃x ∈ K, (x, y) ∈ E ⇐⇒ ∃x ∈ K, f (x) = y ⇐⇒ y ∈ f (K)
We now show that the fiber f −1 ({y}) is compact for all y ∈ Y . For fixed y ∈ Y , regard f −1 ({y}) as a
topological space with the subspace topology of X, and let Z be an arbitrary topological space. Consider the
space f −1 ({y}) × Z, and let π1 : f −1 ({y}) × Z → Z denote the projection map onto Z. Let K ⊂ f −1 ({y}) × Z be
a closed subset. Define a map g : Z → Y, g(z) := y. Clearly, g is continuous. Let π2 : f −1 ({y}) ×Y Z → Z denote
the projection mapping of the fiber product into Z. Since K ⊂ f −1 ({y}) × Z is closed, K ∩ (f −1 ({y}) ×Y Z) =
K ⊂ f −1 ({y}) ×Y Z is closed. Hence, π1 (K) = π2 (K), which is closed subset of Z by our hypothesis that f is
proper. Since Z was an arbitrary topological space, by our previous result, we conclude that f −1 ({y}) is compact.
We now prove the ⇐ direction. Let Y 0 be a topological space and g : Y 0 → Y be a continuous map. Denote
the restriction of the projection map X × Y 0 → Y 0 to X ×Y Y 0 by π : X ×Y Y 0 → Y 0 . Let K ⊂ X ×Y Y 0
be closed, let (A, ≤) be a filtered set, and let A → Y 0 , α 7→ yα0 ∈ π(K) be a net such that yα0 → y 0 ∈ Y 0 .
Then there exists a net xα in X such that xα × yα0 is a net in K. By construction f (xα ) = g(yα0 ). Since g is
continuous, g(yα0 ) → g(y 0 ) = y. Hence, f (xα ) → y. Since X is closed in X and f : X → Y is a closed mapping
by hypothesis, we have that f (X) ⊂ Y is a closed subset. y ∈ f (X) = f (X), hence there exists x ∈ X, f (x) = y,
and (x, y 0 ) ∈ X ×Y Y 0 .
Consider the sets {xβ : β ≥ α}, where α ∈ A is fixed. Since f is closed, f ({xβ : β ≥ α}) is closed in Y ,
hence f ({xβ : β ≥ α}) = f ({xβ : β ≥ α}). Let U be an open neighborhood of y. Since f (xβ ) → y, there exists
γ ∈ A, such that for all β ≥ γ, f (xβ ) ∈ U . For any α, γ ∈ A, there exists λ ∈ A, with α, γ ≤ λ. Then
∀β ≥ λ, f (xβ ) ∈ U ⇒ U ∩ f ({xβ : β ≥ α}) 6= ∅, hence y ∈ f ({xβ : β ≥ α}) = f ({xβ : β ≥ α}), since U was
arbitrary. Using the hypothesis that A is a filtered set, we obtain that
f −1 ({y}) ∩ {xβ : β ≥ α} =
6 ∅,
2
∀α ∈ A
We define a net A → X, where α 7→ x̃α ∈ {xβ : β ≥ α} ∩ f −1 ({y}). Since f −1 ({y}) is compact, this net has
a convergent subnet
B → X, η 7→ x̃η → x ∈ f −1 ({y})
I claim that x is an accumulation point of the net xα . Let U 3 x be an open neighborhood and α ∈ A.
∃α0 ∈ A, ∀η ∈ B, η ≥ α0 , x̃η ∈ U . By construction, x̃η ∈ {xγ : γ ≥ η}∩f −1 ({y}). Since U is an open neighborhood
of x̃η , U ∩ {xγ : γ ≥ η} =
6 ∅ ⇒ ∃γ ∈ A, γ ≥ α, xγ ∈ U .
Equivalently, there exists subnet xαγ → x, and we take the corresponding subnet yα0 γ of yα0 . Therefore,
xαγ × yα0 γ → x × y 0 . Then (x, y 0 ) ∈ K = K, since K is closed. Therefore, x, y 0 ∈ K ⇒ π(x, y 0 ) = y 0 ∈ π(K).
Since yα0 was an arbitrary net, we see that π(K) = π(K) or equivalently, π(K) is closed in Y 0 .
We now give a proof of the ⇐ direction, which does not use the concept of nets.
a ∈ Y 0 \ π(K) ⇐⇒ ∀x ∈ X, (x, a) ∈
/ K ⇐⇒ (X × {a}) ∩ K = ∅
Since f is a closed map and g is continuous, g −1 (Y \ f (X)) =: U is an open subset of Y 0 . Hence, (X × U ) ∩ (X ×Y
Y 0 ) = ∅. If a ∈ U , then we’re done, since U ⊂ π(K). Otherwise, f −1 ({g(a)}) is a nonempty compact subset of
X. Since (X ×Y Y 0 ) \ K is open in X ×Y Y 0 , for each (x, a) ∈
/ K, there exist open neighborhoods Vx ⊂ X and
Vx,a ⊂ Y 0 of x and a, respectively, such that
(Vx × Vx,a ) ∩ X ×Y Y 0 ∩ K = ∅
Since Vx : x ∈ f −1 ({g(a)}) is an open cover of f −1 ({g(a)}), by compactness, there exists a finite subcover
Vx1 , · · · , Vxn , where f (x1 ) = · · · = f (xn ) = g(a). Hence,
Va :=
n
\
Vxj ,a ⊂ Y 0
j=1
is an open neighborhood of a and V :=
Sn
j=1
Vxj is an open subset of X such that
(V × Va ) ∩ X ×Y Y 0 ⊂ (X ×Y Y 0 ) \ K
I claim that Va ⊂ Y 0 \ π(K). Indeed,
y 0 ∈ π(K) =⇒ ∃x ∈ X, (x, y 0 ) ∈ K =⇒ (x, y 0 ) ∈
/ (V × Va ) ∩ X ×Y Y 0
3