E.1
Normality of Linear Continua
is normal in the order topology.
X
Every linear continuum
Theorem E.1.
For if
and no smallest element.
form a new ordered set
(,1)
by taking the disjoint union of
Y
X, and
and
X.
to be less than every element of
(,1)
declaring every element of
The
X
is a linear continuum with no largest or smallest. ,Sinc
Y
ordered set
but no largest, we can
x0
has a smallest
X
has no largest element
X
It suffices to consider the case where
Proof.
implies normality of
Y
Y, normality of
is a closed subspace of
X.
The other cases are similar.
We follow the outline of Exercise 8
hs no largest or smallest.
X
So,suppose
of §32.
Step 1.
component of
c
ard
c'
Let
h
C
a nonempty closed subset of
(-p,c)
or
let us take the union
U
(c,+cm)
has the form
X - C
are points of
Given a point
or
X - C,
x
and lie in
X
that contain
We show that
U
has one of the given forms, and that
components of
Let
a
about
because
or
is not a point of
a
x
U
C.
then the union
and lies in
Similarly, if
is not
b
according as the set
b = +O
If
X - C.
+ a,
a
-
,
(d,e) U (a
of
X- C
X-
that contains
a
at
,bL )
has an
is a point of
(d,e)
for some
then
b
.
is an open interval
This contradicts the definition of
must be a point of
[The form
C.
a.
We
(-a$,+0a)
is nonempty.]
C
It now follows that, because the end points of
no larger subset of
we show
fb )
Then there is an open interval
is of one of the specified forms.
i0 not possible, since
has a lower
BanJ
This open interval contains
C.
;
a = inf a
conclude that
is one of the
U
according as the set
U = (a,b).
Then
disjoint from
that contains
a = -c,
or
b = sup b
Let
upper bound or not.
Suppose that
is connected.
U
X - C.
a = inf a
bound or not.
where
of all open intervals
Tlen
X - C.
of
(a~ ,b, )
(c,c'),
C.
of
x
We show that each
X.
can be connected.
C
x.
lus
+ oor in
C,
U
are
U
must be the component
C.
E.2
Step
2.
component
Let
W
of
dw.
and
B,
B
We show that
x
D.
Let
s:ow that
dW
f
I
I
D,
D
then
I
W.
is a limit point of
D,
is not in
A,
intersect both
A
S:ppose
V
or
W = (b,a),
Let
A
a0< b.
and
AU B,
(a0 )b0 )
D.
(a0 ,b]
[ELOb].
a
of
A
A,
Then
UA
Then
X - A UB
X - AUB;
contains a point of
b0
UA
UB
ard
I
we
contains the point.
W,
which has one
and
be B.
i
We show that
V
Because
cannot
b
of
B; assume for
must contain the interval
A [a,b].
AU B
Tlen
aC
lies in
Let
b0
be the
A.
B
and
b0' aO.
The
because its end points lie in
can be connected.
Hence
as such, it contains a
(a0 ,b0 )
oint of
D.
D, contrary to construction.
X- D
are open sets of
X - D
be the union of all components of
UB
limit points.
(d
lies in
be the union of all components of
and let
A
is not a limit point
and a point
V
Step 4. By Step 1, the components of
Let
x
ae A
does not intersect
contains no point of
is one of the components of
V
has no
If
X - D.
Being connected,
no larger subset of
Hence
lies in both
B.
set
B
x
W = (b,a).
be the supremum of the set
infemum of the set
interval
and
Thb
be the set of all the
that is disjoint from A;
Where
be a component of
a< b.
a0
A
it can intersect at most one set of the form
contains a point
convenience that
[a,b].
V
A
D
and show that
a L.
Let
D
t1hen
It follows that
and at most one set of the form
Step 3.
Let
intersects the corresponding set
W = (a,b)
For each
is closed and discrete.
be an open interval about x
is disjoint from
W - (a,b)
d . in
contains at most two points of
of the forms
I
x
X.
that is an open interval with one end point in
(which is not possible).
We suppose that
of
be disjoint closed sets in
choose a point
We show that if
and
B
X - A VB
and the other in
points
A
X.
that intersect
X - D
are disjoint open sets containing
A
that intersect
and
B,
B.
respectively.
a