K.1
Compactly generated spaces
A space
Definition.
the following condition:
X
is said to be compactly generated if it satisfies
A set
for each compact subspace
C
A
of
is open in
X
if
An C
is open in C
X.
Said differently, a space is compactly generated if its topology is coherent
with the collection of compact subspaces of
of coherent topologies.)
X.
(See Notes G for a discussion
Many spaces are compactly generated; for instance,
locally compact spaces are compactly generated, and so are first-countable spaces.
(See Lemma 46.3.)
Compactly generated spaces are useful when studying various topologies
on the space
(X,Y)
of continuous functions
in other contexts as well.
f : X ->Y,
but they occur
Here we explore their relation to proper maps
And to perfect maps.
Definition.
subspace
C
of
A map
f : X--.Y
Y, the subspace
Roughly speaking,
f
is said to be proper if for every compact
-1
f (C) of X is compact.
is proper if it does not collapse any subset of
that runs off to infinity onto a compact subspact of
Y,
X
which does not run off
to infinity.
Theorem K.1.
Let
f: X---Y
generated Hausdorff space, and if
Proof. Let
A
of
Y.
f(A)n C
f
(C)
A
is proper,
X.
then
To show
is closed in
C
If
f
Y
is a compactly
is a closed map.
f(A) is closed, we
for each compact subspace
Now
f(A)(\C
The space
f
be a closed set in
need only to show that
C
be a continuous map.
f
(C)
=
is compact because
is also compact.
and is therefore closed in
f(f l(c)C A).
f
is proper, so its closed subspace
Thie image of this set under
Y. i
f
is compact,
K.2.
Corollary K.2.
is proper, and
Let
is compactly generated Hausdorff, then
Y
whose image is a closed subspace of
Example 1.
f: [0,2)
Let
f
f
If
be continuous and injective.
f : X-->Y
is an imbedding.
Y.
be given by the equation
-­
>R2
f(t) = (cos t,sin t).
f is continuous and injective, and its image is the unit circle, which
Then
is closed in tR2 .
Ard
circle is not compact.
A map
Definition.
is not proper; the inverse
f
However,
f
image of the unit
is not an imbedding.
isE said to be perfect if it is continuous,
f: X->Y
closed, and surjective, and if
f
is compact for each
(fy})
yY.
Said differently, a perfect map is a closed quotient map such that the inverse
Perfect maps have many special properties.
image of each point is compact.
f : X --Y
For instance, if
is perfect and
is compact, then
Y
the same result holds if "compact" is replaced by "paracompact."
other hand, many "niceness" properties of
X
is compact;
X
On the
(such as the Hausdorff condition,
regularity, local compactness, and second-countability, as well as the condition
of being paracompact Hausdorff) are preserved by perfect maps.
(See Exercise 12
of §26, Exercise 7 of §31, and Exercise 8 of §41.)
The relation between perfect maps and proper maps is given in the following
theorem:
If
be continuous and surjective.
f
Hausdorff, then
Y;
let
4
is proper, and if Y
f
f
is a perfect map.
beian open cover of
fl (C).
Let
map, there is a neighborhood
.
many such neighborhoods
W
of
y
.
y
C,
Because
such that
(See Exercise 6 of §31.)
W;
ke a compact subspace
C
Given
can be covered by finitely many elements of
same elements of
is compactly generated
is perfect.
Proof. -Suppose
of
Conversely, let f: X-Y
Every perfect map is proper.
Theorem K.3.
f- (W)
tie set
f
(y})
is a closed
is covered by these
We can cover
then thbeir inverseimages
p
cover
C by finitely
neighborhoods
-Wt.
1many
then
fsuch
(C).
K.3
Now we supose that
Y
f
is continuous, surjective, and proper, and that
T1-ie fact that
is compactly generated Hausdorff.
immediate, since
fy)
is compact.
Tile fact that
f- (y)
f
is compact is
is closed follows
from Theorem K.1. t1
It is an, interesting fact that if X
is not compactly generated, it may
be given a (finer) topology that is compactly generated, and has exactly
the same collection of compact subspaces:
Theorem K.4.
Let
XT
be a space with underlying set
There is a unique topology
(i)
If
D
XC ,
on
is a subset of
it inherits from
from
C
XT ,
or if
X,
X,
D
finer than
and if
XC
Proof.
Let
{C~
that each space
CW
Suppose
D
D
C
XC
Now suppose
D
X, finer than
T,
such
and the topology of
Xc
is coherent
C
X
i : XC -- XT
XT .
is continuous,
U
nD
Because
if and only if
D
C'
is open in
D
XT
if and only if it is
nD
is open in
XC .
D
Therefore a set
U
is
be any topology satisfying the conditions
is a compact subspace of
and only if it is open in
U
for each compact subspace
is compactly generated, a
XT .
is compact in
For by definition,
XC,
U
is a compact subspace of
D
XC.
Then the previous paragraph applies.
is compactly generated.
if and only if
By (i),
Then
is compact in the topology it inherits from
To prove uniqueness, let
XC,.
XT .
; and as just noted, each space
D of XT. But D is a compact subspace of
a compact subspace of XC .
cf
In
XC.
the topology it inherits from
of the theorem.
XT.
.
Because the identity map
The space
on
is compact in the topology it inherits from
is a subspace of
irn XC,
is compact in the topology
are the same.
C
in this topology, it is one of the spaces
XC
such that:
be the family of compact subspaces of
is a subspace of
with the subspaces
open in
T.
is compactly generated.
view of Theorem G.4, there is a topology
Ca
and topology
is compact in the topology it inherits
then these two topologies on
(ii)
D
T,
X
set
U
is open
for each compact subspace
XC,
U
if and only if it
is open in
XC,
if
D
K.4.
The class of compactly generated spaces is an interesting one to explore.
Like the class of normal spaces, it is not closed under the operations of taking
subspaces or products.
We shall show that if
is not compactly generated.
J
[0,1]
J
is uncountable, then
It follows that the subspace
is not compactly generated, although [0 ,1 ]J
compactly generated.
(0,1)
~J
of
is compact and thus
It also follows that an arbitrary product of compactly
generated spaces need not be compactly generated. (The same is true for
finite products, but the required example is more complicated.
Example 2.
If
is uncountable, then IRJ
J
See [D], p.249.)
is not compactly generated.
(This example is adapted from [Wd].)
Given
x
n>1,
A
be the set of all points x of IRJ
such that
n
values of
, and x = n for all other values of (.
let
= 0 for at most
n
We show that each set
I!t
p
A
is closed in tR
be a point of [RJ
not in
AD .
If
44
p
O,n) for some P,
-1
let
U
be a neighborhood of
neighborhood of
since
p
disjoint from
U
=IR
from
An.
If
p
= 0
J0
'(J 0o
for
0 or
pC
_ An, there must be a finite set
p
n elements such that
and
pi not containing
n;
0,nJ
then 7i3 (U)
for all 4<,
Setting
U=
(-1,1)
TrU£
of
for
p
J-0
X
disjoint
A
C
is a compact subspace of J
IR
only finitely many of the sets
An .
Since
C
then
R.
Then
many sets
C
An.
lies in
Since
J
1-IA
An
; we show that
JO
of
does not intersect [|I~ if
all but finitely many
infinitely many
x
iWF
I
is uncountable, the map
must map some infinite subset
J
n>N.
x )must
be less than or equal to
1
] - I.
ntersects only finitely
o(-- n,
For if
N
intersects
1T,(C);
[-n,n
of
to a single integer
are greater than
C
is compact, so is
therefore the latter is contained in some closed interval
that
then
of indices containing more than
otherwise, we obtain a neighborhood
We now show that if
of
is a
x
while if
N.
J
N.
belongs to
x
into
4
It follows
An,
belongs to
then
T-Ig,
K.5
Let
J
in IR ,
T
be the union of the sets
but that
T
C
is closed in
It is easy to see that
of
T
that is not in
T.
T
ard
x
C
.
We show that
RJ, for
Given a basis element
for which
= n otherwise (where n
U
U
# R.
subspace of
the form
R
Fc;r in this case
An n C; and since
An
T NC
is not closed
O
C
containing
If we set
C
if
of JIR
is a limit point
0,
C
let
J0
At= 0
is the number of elements in
we obtain a point x of T that lies in
-U
.
It is also easy to see that TnC is closed in
J .
T
for every compact subspace
is not closed in
be the (finite) set of indices od
for cIJ0
A
JO) ,
is a compact
is the union of finitely many sets of
is closed in
RJ ,
the set
Ann C
is closed in
C.