1st December 2004
Munkres
§
24
Ex. 24.2 (Morten Poulsen).
Let
f
:
S
1
→
R
be a continuous map. Deﬁne
g
:
S
1
→
R
by
g
(
s
) =
f
(
s
)

f
(

s
). Clearly
g
is continuous. Furthermore
g
(
s
) =
f
(
s
)

f
(

s
) =

(
f
(

s
)

f
(
s
)) =

g
(

s
)
,
i.e.
g
is an odd map. By the Intermediate Value Theorem there exists
s
0
∈
S
1
such that
g
(
s
0
) = 0,
i.e.
f
(
s
0
) =
f
(

s
0
).
This result is also known as the BorsukUlam theorem in dimension one. Thus there are no
injective continuous maps
S
1
→
R
, hence
S
1
is not homeomorphic to a subspace of
R
, which is
no surprise.
Ex. 24.4.
[1,
§
17]. Suppose that
X
is a linearly ordered set that is not a linear continuum. Then
there are nonempty, proper, clopen subsets of
X
:
•
If (
x, y
) =
∅
for some points
x < y
then (
∞
, x
] = (
∞
, y
) is clopen and
6
=
∅
, X
.
•
If
A
⊂
X
is a nonempty subset bounded from above which has no least upper bound then
the set of upper bounds
B
=
T
a
∈
A
[
a,
∞
) =
S
b
∈
B
(
b,
∞
) is clopen and
6
=
∅
, X
.
Therefore
X
is not connected [2,
§
23].
Ex. 24.8 (Morten Poulsen).
(a)
.
Theorem 1.
The product of an arbitrary collection of path connected spaces is path connected.
Proof.
Let
{
A
j
}
j
∈
J
be a collection of path connected spaces. Let
x
= (
x
j
)
j
∈
J
and
y
= (
y
j
)
j
∈
J
be
two points in
Q
j
∈
J
A
j
For each
j
∈
J
there exists a path
γ
j
: [0
,
1]
→
A
j
between
x
j
and
y
j
, since
A
j
is path connected
for all
j
. Now the map
γ
: [0
,
1]
→
Q
j
∈
J
A
j
deﬁned by
γ
(
t
) = (
γ
j
(
t
))
j
∈
J
is a path between
x
and
y
, hence the product is path connected.
±
(b)
.
This is not true in general: The set
S
=
{
x
×
sin(
x

1

0
< x < π

1
}
is path connected, but
S
=
S
∪
(
{
0
} ×
[

1
,
1]) is not path connected, c.f. example 24.7.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Brown
 Topology, Morten Poulsen

Click to edit the document details