There is a bijection between [0
,
1] and [0,1]2. Why? There exists an injection [0,1]→[0,1]2
1]
2
→
[0
,
1].
The details of this second map are as follows.
Write
x
=
∑
∞
k
=0
x
k
10

k
with 0
≤
x
k
≤
9 and the decimal expansion not terminating in an infinite
sequence of 9s. If
x
= 1 then
x
0
= 1 and
x
k
= 0 for
k >
0.
y
=
∑
∞
k
=0
y
k
10

k
with the same
properties as
x
above.
Define the function
h
: [0
,
1]
2
→
[0
,
1] by
h
(
x, y
) = 0
.x
0
y
0
x
1
y
1
. . .
=
∞
k
=0
x
k
10

(2
k
+1)
+
∞
k
=0
y
k
10

(2
k
+2)
.
h
: [0
,
1]
2
→
[0
,
1] is not onto, but it can be easily be verified that
h
is onetoone. Then
by the SchroederBernstein Theorem, there exists a bijection [0
,
1]
→
[0
,
1]
2
.
The Peano
Curve is a function
f
: [0
,
1]
→
[0
,
1]
2
which is onto and continuous.
(Note: There is no
homeomorphism [0
,
1]
→
[0
,
1]
2
.)
Lemma 6.5.
Suppose
X
is compact and
f
:
X
→
Y
is continuous, onto and onetoone.
Then
f
is a homeomorphism.
Proof.
We can define
g
:
Y
→
X
with
g
(
y
) =
x
if
f
(
x
) =
y
. Given an open
V
⊂
X
, we need
to check that
g

1
(
V
) is open in
Y
.
K
=
X
\
V
is closed in
X
, hence it is compact. Therefore,
f
(
K
) is also compact.
f
(
K
) =
g

1
(
K
). Then
g

1
(
V
) =
g

1
(
X
\
K
) =
g

1
(
X
)
\
g

1
(
K
)
which is open in
Y
.
Definition 6.6.
A (topological or) metric space
X
is
pathconnected
if for any
x, y
∈
X
there exists a path connecting them, i.e.,
∃
γ
: [0
,
1]
→
X
, which is continuous with
γ
(0) =
x
and
γ
(1) =
y
.
Clearly,
R
k
and [
a, b
] are connected.
More generally, if
K
⊂
R
is
convex
, it is path
connected.
K
is convex if for all
x, y
∈
K
and for all
t
∈
[0
,
1],
γ
(
t
) =
ty
+ (1

t
)
x
∈
K
.
Definition 6.7.
A set
K
⊂
X
is called
clopen
if it is closed and open in
X
.
Definition 6.8.
X
is
connected
if the only clopen sets in
X
are
∅
and
X
.
Equivalently,
X
is connected if and only if for any partition
X
=
V
1
∪
V
2
of
X
into two open
and disjoint sets, one of them is
∅
. For instance,
Q
=
{
x
∈
Q
:
x <
√
2
} ∪ {
x
∈
Q
:
x >
√
2
}
and thus
Q
is not connected.
29
30
Lecture 12: October 06
Fact 6.9.
For
a < b
∈
R
,
[
a, b
]
is connected.
Proof.
Suppose
K
⊂
[
a, b
] is clopen in [
a, b
] and
K
=
∅
,
K
= [
a, b
]. If
a
∈
K
, let
s
= sup
K
⊂
[
a, b
]. Then
s
∈
K
because
K
is closed. We have three cases:
(1) Suppose
s
=
a
.
This is impossible because we know, since
K
is open, that there
exists
>
0 such that [
a, a
+ )
⊂
K
.
(2) Suppose
s < b
.
This is impossible because we know, since
K
is open, that there
exists
>
0 such that [
s, s
+ )
⊂
K
.
(3) Suppose
s
=
b
. This implies that
b
∈
K
. To get a contradiction, we examine sup
K
c
.
(a) Suppose sup
K
c
< b
. We apply the argument in (2) and reach a contradiction.
(b) Suppose sup
K
c
=
b
. Then
b
∈
K
c
. But
b
∈
K
, so we have a contradiction.
Theorem 6.10.
Suppose that
X
is pathconnected. Then
X
is connected.
Proof.
Assume
X
is not connected.
Suppose
X
=
V
1
∪
V
2
with
V
1
, V
2
open, disjoint and
nonempty. Let
x
∈
V
1
,
y
∈
V
2
. We find
γ
: [0
,
1]
→
X
with
γ
continuous and
γ
(0) =
x
and
γ
(1) =
y
. Then [0
,
1] =
γ

1
(
V
1
)
∪
γ

1
(
V
2
) is a disjoint union of open sets and nonempty
since 0
∈
γ

1
(
V
1
) and 1
∈
γ

1
(
V
2
). This is a contradiction since [0
,
1] is connected (meaning
one of
γ

1
(
V
1
) and
γ

1
(
V
2
) has to be empty.)
Fact 6.11.
[0
,
1]
and the circle
C
=
{
(
x, y
)
∈
R
2

x
2
+
y
2
= 1
}
are not homeomorphic.
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