C.1
The Long Line
We follow the outline of Exercise 12 of §24.
Let
denote the set
L
denote the smallest element of
Let o
Lemma C.1.
[o 0 X 0, ckX 0]
interval
S
of
has the order type of
L
0 in 5
S_.
Sppose the lemma holds for all
We show it holds for 3.
.
4<
[0,1].
i: the immediate successor
Proof. Note that the proof is trivial if oL
of
Tren the
o0.
SLddifferent from
bEx a point of
(
the order topology.
L
Give
.
Let
in the dictionary order.
Stx [0,1),
If
has an immediate predecessor c1' the proof is easy. The interval
[dOx 0, lX 0] of L has the order type of [0,1] by hypothesis.
Tlhe interval
l ; .
[
has the order type of
of
L
of
[0,2],
=
<[0, 1)) L I
equals (
and also of
[0,1],
[],2]
[0,1]
order type of
<O]
,
KO
Their union has the
[1,2].
which of course has the order type
[0,1].
an increasing sequence
3 haIs no immediate predecessor, there is
If
whose supremum is P . Assumle o
of points of S
V1' 2''.
convenience. We show that for each i the interval [do i 0,
has the order type of
[0,1]
of
The interval
[0,1].
[0,1] by hypothesis; if
order type of
c
sc: it
[<
0, aOi+l X 0]
has the
ix0, i+l
[
[c,l], which of course has the order type of
O]
[0,1].
Finally, we note that the interval
J = [ OX 0, 0)
of
L
can be written as the union
[0
X
of intervals of
X
0 , liX O]u [
L.
Of
2
2X 0] ... U [i< O,
i+l
X
0] J...
There is an order-preserving correspondence of this
union with the union
[0,1]U [1,2]U ...
[i, i+l]d ...
of intervals of IR. The latter union equals
type of
[0,1).
When we adjoin the point P
with the order type of
[0,1].
]
L
corresponds to the real number
iX 0
under the order-preserving bijection, then
has the order type of
fcr
> 0
,i+l 0] of
[0,+oa), which has the order
O
to
J,
we obtain a set
C.2
Let
Definition.
L - 9 0 w 0} of
be the subspace
L'
L; it is called
the Long Line.
every
The long line is a path-connected linear continuum,
Theorem C.2.
point of which has a neighborhood homeomorphic to an open interval of
[R.
It is not metrizable.
Proof.
C
of
(
w
of
R.
Then
x < (x0.
to that
S
L
be a point of
x
Let
with
x
lies in the open interval
of
L,
which has the order type of the open interval
qT-e fact that
L'
i& a linear continuum follows from Ex. 6 of §24.
x 0)
o 0,
X
(g 0
open intervals
0,AXO)
co0 x
since they have the point
Now let
R
=
/
L'
½ in common,
of
L
R
countable) subcovering.
then
S.
[a X
To
show
0)
R
The set
If
Se.
with
A i>L
It;
has no finite (or even
is limit point compact, it suffices
S
in
R
has a limit point.
S
And
hs an upper
is the immediate successor of this upper bound,
[Gdx O
X 0]
of
Since
L'.
is a linear continuum, this interval is compact; therefore
point. [l
We show that the
L'.
of first coordinates of points of
is a subset of the interval
S
in
is limit point compact but not compact.
to show that every countably infinite set
bound in
0
is not compact follows from the fact that the covering
by the open sets
this is easy:
is path connected.
L'
is not metrizable, so neither is
R
The fact that
of
L, each of which is path connected;
be the immediate successor of
[ [X 0, +)
follows that
of
(0,1)
is the union of the
L'
The result of the preceding paragraph shows that
ray
Choose an element
LOX 0.
x
S
L'
has a limit