F.l
The separation axioms
We give two examples of spaces that satisfy a given separation
axiom but not the next stronger one.
Te first
is a familiar space,
and the second is not.
Teorem F.1. If
not normal.
Proof.
completel
of R J
is uncountable,
is of course completely regular, being a product of
regular spaces.
Le't
X = Z+ ; since
it suffices to show that
notation for the elements of
Given a finite subset
U(x,B)
X
product
in
B. Then
FU~,
where
otherwise.
of
J
U(x,B)
x
i
n
X;
irldeed, it is the cartesian
x
Pn
Pn
U(x,B)
In) consists
is closed, for if
disjoint from
U(y,E),
is in
P , then
if
is in
We show that
and
U
be the subset of
X
of
where
B =
J
X,
consisting
i dfferent from
is not in
,"
= 2+
form a basis for
J.
n,
(This of course
of uncountably many elements of
y
)
J.)
Pn, then there is an integer
such that
,jiC,
y(4 ) = y(
contains
y
) = i.
and is
Pn
Furthermore, if n
U
B
= x(
7.
such that for each
and distinct indices
L!t
n, let
in
y(od
let
is open in
(i) consists, of at most one element of
The basis element
x
such that
X,
X
-1
implies that
The set
x :J-47
of
of
is a one-point set fr
Given a positive integer
x
y
since the one-point sets form a basis for
the set
WE! shall use functional
and given a point
It is immediate that the sets
of those maps
is a closed subspace
X rather than tuple notation.
B
U
X
i not normal.
be the set of all those elements
for all a
J is completely regular, but
The proof follows the outline given in Exercise 9 of §32.
RJ
The space
J
P ,
and
U
m,
then
Pn
and
Pm
are disjoint.
x maps uncountably many elements of
it maps at most one element of
V be open sets of
and
v
X
containing
are not disjoint.
J
J
to
to
n.
P1
ard
It follows that
For if x
n;
P2
X
while
respectively.
is not normal.
F.2
Step 1.
sequence
We define a sequence
xl, x 2 ,...
1'
of points of
X,
2'",...
of elements of
ar;d a sequence
nl < n
J,
a
< ..
2
of positive integers, inductively as follows:
Let
subset
x(A) = 1
B1
of
elements of
J
B1
for all o(.
such that
Let
xk
U(x 1, B 1)
Bk
and
in
x+l
belongs to
U.
Index the
are given, and that
dj
is defined for
of
=
,j |1
X
by setting
P1 .
= 1
Cloose
1j<nk ,
l
for
Bk .
Bk+l
Index the elements of
B +- -Bi
=f
and
for all other ct.
so that
Without loss of generality we can choose
contains
choose a finite nonempty
U.
nk
XI+l(o)
x
Pi;
is contained in
:+l(d;) = j
Then
is in
denote the set
Bk
Define a point
xl
so that
Now suppose that
j = 1,...,n k .
Then
Bk+l - Bk
U(xk+l,Bk+l)
B+
is contained
so that it properly
1
so that
(j nk.ij
inkl
.
By induction (actuall.y,recursive definition), we have defined
8nd
ni1
y
Since
nk
and a(.i
for all i.
Step 2.
Then
x
Now, define a point
belongs to
C
intersects
C
of
X
by setting
y(j")
=
j
for
y( A)
=
2
for all other c.
Choose
C
P2.
is finite, it contains
so that
y
contains no
U(xk+l,Bk+l),
Let us define a point
(j
all
so that
U(y,c)
is contained in
dj for only finitely many
for which
j-in
.
sO that the sets
z
j ,
of
X
U
ad
choose
We shall show that
V
z( j)
=
j
for
1
z(j)
=
1
for
nk< j $n+l,
z(cd)
=
2
j n k
.
U(y,C)
are not disjoint.
(cleverly!) by setting
for all other
j;
V.
and
F 3
Then
to
z(ij)
equals
U(xk+l,Bk+l).
el in
xk+l(i)
z belongs to
certainly true that
z(,)
if
A
Proof.
z( )
= y(ot)
for
if ot is one of the indices
z( j) = j = y(Olj).
It is
j:., for
And it is true that
is not one of the indices o(j;
for in that case
[-1,0]
C
n,k
There is a space that is regular but not completely regular.
The proof follows the outline given in Exercise 11 of §33.
Step 1.
let
we show that
z belongs
= 2 = y(]). Theorem F.2.
m
sc that
U(y,C); our result is then proved.
z(o ) = y( o)
j<n k, so that
z(o ) = y( oL)
l<jznk1,
On the other hand,
C, so that
in that case
for
Given an even integer
in the plane.
m, Let
Lm
denote the line segment
And given an odd integer
n, and an integer
k> 2,
denote the union of the line segments
(r + (k-l)/k)
-1,0]
(rL
(k-l)/k)X [-1,O]
(n- (k-i)/k),C-l1O0
and the semicircle
fx
in the plane.
We call C
ar
n, k
Finally, we let
m,
y I(x-n)2 + y2 = (k-1)2/k2
X
ard the arches
'.arch"
y> O0
and we call: L
m
be the union of the pillars
L,
Cn k
n
for all odd integers
along with two additional
points
infinity."
n
For each odd
Pn,k
it is the "peak" of the arch
a
ard
and each
=
n
C
a "pillar."
for all even integers
ar:d all integers
k.
k_>2,
b, which we call the "points at
k>2,
we let
Pn,k
be the point
(k-l)/k ;
See the accompanying figure.
ndk
Pl
ard
F.4
We now topologize
X
in a most unusual fashion.
We take as basis
elements all sets of the following five types:
fp3,
(i) Each one-point set
of the arches
Cnl
where
p
is a point lying on any one
that is different from the peak
(ii) The set formed from one of the sets
Pn,k
of this arch.
Cn,k
k
by deleting finitely
many points.
(iii)
Fcr each even integer
y [-1,0],
the intersection of
, m+
(m -
)
(iv)
xxy
of
m,
X
for which
x
xXy
0 <E1, and each
with the horizontal open line segment
of
X
m;
te union of
m,
the union of
a}
ar;d the set of points
m.
(X') For each even integer
points
with
y.
Fcr each even integer
X
each
for which
b}
ard the set of
x>m.
The basis elements of tpe (ii) are the neighborhoods of the peaks; those
of type (iii) are the neighborhoods of points lying on the pillars; and those
of types (iv) and (v) are the neighborhoods of the points at infinity.
It is
easy (but boring) to check the conditions for a basis; we leave it to you.
Each of the arches
C
is an open set of X.
n,k
We shall call the space X
"Thomas' arches,"
because it was invented
by the topologist John Thomas.
Step 2. It is trivial to check that
X
is a T-space; given two points,
each has a neighborhood that excludes the other.
let
p
be a point of
X;
and let
U
To check regularity,
be a basis element containing
We consider several cases, showing there is a neighborhood
that
is a basis element of types (i), (ii), or (iii), then
we are finished.
point
a
the point
x
of
p
such
VcU.
If U
and
V
p.
y
If p
of
So suppose that
along with those points
a, then we let
X
for which
V
(ii), or (iii) containing
p
?Te argument when
is of type (iv), consisting of the
of
consist of
x< m- 2. Then
is some other point of
finished.
xX y
U
V
X
for which
the point
= VU TL-_
a
2
xm. If p
and lying in
U;
then
is
along with those points
which lies in U.
U, there is a basis element
U
U = U,
V = V
is of type (v) is similar.
V
of type (i),
ar;d we are
F.5
Step 3.
X
nk, let
f
which the value of
f
The set
open sets
is countable:
Sn, k
Tlen the set
Cn.
n,k'
Cn k.
of
points
many
countably
Cnki
the value of
d
-
with
f
Cm+2
JR x d
= c.
O
Sn
k.
Therefore
sulch that the horizontal
This means that for each
cm
b
the point where the line
IR x dl
cm
at the points
f
are equal.
bk
and
d
is countable.
S
We assert that the values of
To prove this fact, set
ak
f(Pn) k
at the peak of the arch.
let
m,
Now for each even integer
and
Let
at the points where the arch intersects this
f
horizontal line equals the value of
intersects the pillar L .
at the
is countable.
Snk
Tlus
d3 intersects none of the sets
arch
f
for
Each of these open sets contains all but
It follows that the union of all the sets
R
k
Hence their intersection contains all but
finitely many points of
line
Cn
X, since it is the intersection of the
(c) is a G-set in
-1
1
1
f ((c - n , c + )).
we may choose a real number
of the arch
p
is different from the value of
p
at
Of the arch.
Pnk
peak
be the set of points
Sk
f
f(a) = f(b).
then
f:X--O[0,1],
is any continuous function
Given
Indeed, we show that if
is not completely regular.
.
n = m+],
C
consider the arch
,1
and let
denote the points of intersection of this arch with the line
(For convenience, let
k
bk
converges to c +2 .
to
f(cm )
f(a)
k
We conclude that
=
then implies that
f(pnk)
=
cm
converges to the point
as
m
converges
But by construction,
f(bk)
a
as;
f
and
m goes to -wo,
f
··
I·
>
"
I
L,,
LA-n
it
~ft4/
I
C
are all equal .
cm
at the points
It follows from continuity of
+.
.
f(ak )
f(cm ) = f(cm+2)4
It follows that the values of
goes to
f
f(cm+2).
converges to
f(bk )
and
Continuity of
wile the sequence
c ,
converges to
ak
increases, the sequence
as
Then
be the one with smaller x-coordinate.)
ak
It" 4 7
cm
that
converges to
f(a) = f(b).
But
b
a