J.1
The Pruifer Manifold.
;The so-called Prufer mnanifold is a space that is locally 2-euclidean
and Hausdorff, but not normal.
In discussing it, we follow the outline of
Exercise 6 on p. 317.
Definition.
Le-t
A
be! the following subspace of
gx, y,)
A=
Given a real number
c , let
c
X
be the set that is the union of
real.
Topologize
X
.
3
TR3:
be the following subspace of
xy
Yc)[ x
Bc=
Let
Be
j x> 0
R2:
0o.
A
and all the spaces
Bc , for
by taking as a basis all sets of the following three
types:
(i) U,
(ii)
where
V,
where
with
(iii)
x
U
is open in
V
A.
is, open in the subspace of
For each open interval
Z>0,
I = (a,b) of
the set
A (I, ) = x,y) '
x
Ac(I,
and
R,
each real
number
c + ax y
= ~Xyc) |-£
x < O and
is called the "Prufer manifold."
at y< b
c+ bx
,
.
Let us sketch what the basis elements of type (iii) look like.
c, I, and
£,
the basis
Uc(I,_)
- +
-VL
- I
=C � a+
I
i
LZt.
t/
$6. a,
(C-~~~~~2_
~~
~
,
Given
is: the union of the two shaded figures
in the figure.
B
~
c,
)v Bc(I, £).= 'JTc4., .), where
B(I,f)
X
consisting of points
0.
and each
The space
B
~
­
J.2
It is easy to check that these sets form a basis for a topology.
The
intersection of a set of type (i) with any other basis element is empty or
is open in
A,
and the intersection of a set of type (ii) with any other basis
element is empty or is open in the subspace
U (I, )
intersection of the sets
c f d; it is empty if c - d
c = d
and
I
intersects
and
and
I
X
U(I',
of
')
B
Finally,the
.
is open in
is disjoint from
A
if
I'; and finally if
I',. it equals the set
Uc(I (I', min(
We show that
x<O
,
')).
is locally 2-euclidean.
For fixed
c,
tIhe set
is a union of basis elements, and therefore is open in
X.
B VA
C
Surprisingly,
R2
! Consider the map
fc
IR2 -X
it is actually homeomorphic to
given
by
The map
fc
= (x,y,c)
fc(x,y)
= (x, c+
carries the subspace of
bijectively ont BC,
x >0
fc(x,y)
for
xy)
R
x
for
0,
x >0.
ccnsisting of points
(x,y) with
x
and it carries the subspace consisting of points with
bijectively onto
A
(Each vertical line
x = x
is carried bijectively
onto itself).
for
To show that f
iS a homeomorphism, we note that we can take as basis
2
RI
all open sets lying the the half-plane x 0, all open sets lying
in the half-plane x >0,
and all open sets of the form
Each of these is mapped by
fe
[The, open set
The map
is pictured in the accompanying figure.
It follows that
open sets
BU A,
X
(- ,
)
is mapped onto
)X
(a,b).
X;
and
U((a,b),).]
is locally 2-euclidean, since it is covered by the
each of which is homeomorphic to
IK-
(- ,
onto one of the basis elements for
conversely.
f
O
TcI
2
.
A
J 3
T1e only case where some care is required
is Hausdorff.
X
WE, show that
is the case where the two distinct points are points of the "edges" of the
Bc
half-spaces
In this case, we need merely choose
of the form (O,yl,c) and (O,y2 ,c).
disjoint intervals
I1
about
12
and
U(I, t)
basis elements.
Choose an open interval
(C,yl,c)
containing both
I = (a,b)
so that
[See the accompanying figure.
(b-a)<(c-d),
d +b<c
so that
A-t
j/'~~~~~
,,,
d<c,
one chooses
+a.]
L
11)
/
I
I
A
/2
TIle proof follows a familiar pattern.
Finally, we show that
X
is not normal.
be the subspace of
X
consisting of all points of the form
L
)
Ud(I,
ar;d
/
/.J
Let
Assuming
)
Then
/
//7
/
Uc(I,
Y2.
and
,9,?e
/
'
Y1
c / d.
where
(O,y2 ,d),
and
).
(for any
are disjoint
is sufficiently small, the basis elements
are disjoint.
A
Y2, respectively; then the
and
Y1
U (I2 , £)
and
Now consider two points of the form
if fE
Bc , then they are
If they belong to the same half-space
.
It is closed in
X;
(O,O,c).
ad it has the discrete topology since each basis
element of type (iii) intersects
in at most a single point.
L
Ore show repeats the argument given in Example 3 of §31, which
If X were normal, then for every subset C
2 is not normal.
isi
R
sw
of X containing
and VC
one could choose disjoint open sets UC
showed that
of
L,
C
and
L- C, respectively.
Letting D
rational coordinates, one defines
:
be the set of points of
(L)-->- (D) by setting
A
having
J.4
o(c)
=
o(0)
= 0
0(L)
=
One shows readily that
DUC,
D.
is injective; then one derives a contradiction from
cardinality considerations, since L is uncountable and D is countable. D