3Connectivity
IB Metric and Topological Spaces
3.2 Path connectivity
The other notion of connectivity is path connectivity. A space is path connected
if we can join any two points with a path. First, we need a definition of a path.
Definition (Path). Let
X
be a topological space, and
x
0
, x
1
∈ X
. Then a
path from
x
0
to
x
1
is a continuous function
γ
: [0
,
1]
7→ X
such that
γ
(0) =
x
0
,
γ(1) = x
1
.
Definition (Path connectivity). A topological space
X
is path connected if for
all points x
0
, x
1
∈ X, there is a path from x
0
to x
1
.
Example.
(i)
(
a, b
)
,
[
a, b
)
,
(
a, b
]
, R
are all path connected (using paths given by linear
functions).
(ii) R
n
is path connected (e.g. γ(t) = tx
1
+ (1 − t)x
0
).
(iii) R
n
\{
0
}
is path-connected for
n >
1 (the paths are either line segments or
bent line segments to get around the hole).
Path connectivity is a stronger condition than connectivity, in the sense that
Proposition. If X is path connected, then X is connected.
Proof.
Let
X
be path connected, and let
f
:
X → {
0
,
1
}
be a continuous function.
We want to show that f is constant.
Let
x, y ∈ X
. By path connectedness, there is a map
γ
: [0
,
1]
→ X
such that
γ
(0) =
x
and
γ
(1) =
y
. Composing with
f
gives a map
f ◦ γ
: [0
,
1]
→ {
0
,
1
}
.
Since [0
,
1] is connected, this must be constant. In particular,
f
(
γ
(0)) =
f
(
γ
(1)),
i.e. f (x) = f(y). Since x, y were arbitrary, we know f is constant.
We can use connectivity to distinguish spaces. Apart from the obvious “
X
is connected while
Y
is not”, we can also try to remove points and see what
happens. We will need the following lemma:
Lemma. Suppose
f
:
X → Y
is a homeomorphism and
A ⊆ X
, then
f|
A
:
A →
f(A) is a homeomorphism.
Proof.
Since
f
is a bijection,
f|
A
is a bijection. If
U ⊆ f
(
A
) is open, then
U
=
f
(
A
)
∩ U
0
for some
U
0
open in
Y
. So
f|
−1
A
(
U
) =
f
−1
(
U
0
)
∩ A
is open in
A
.
So f |
A
is continuous. Similarly, we can show that (f|
A
)
−1
is continuous.
Example. [0
,
1]
6'
(0
,
1). Suppose it were. Let
f
: [0
,
1]
→
(0
,
1) be a homeomor-
phism. Let
A
= (0
,
1]. Then
f|
A
: (0
,
1]
→
(0
,
1)
\ {f
(0)
}
is a homeomorphism.
But (0, 1] is connected while (0, 1) \ {f (0)} is disconnected. Contradiction.
Similarly, [0
,
1)
6'
[0
,
1] and [0
,
1)
6'
(0
,
1). Also,
R
n
6' R
for
n >
1, and
S
1
is
not homeomorphic to any subset of R.
3.2.1 Higher connectivity*
We were able to use path connectivity to determine that
R
is not homeomorphic
to
R
n
for
n >
1. If we want to distinguish general
R
n
and
R
m
, we will need to
generalize the idea of path connectivity to higher connectivity.
To do so, we have to formulate path connectivity in a different way. Recall
that
S
0
=
{−
1
,
1
} ' {
0
,
1
} ⊆ R
, while
D
1
= [
−
1
,
1]
'
[0
,
1]
⊆ R
. Then we
can formulate path connectivity as:
X
is path-connected iff any continuous
f : S
0
→ X extends to a continuous γ : D
1
→ X with γ|
S
0
= f.
This is much more easily generalized:
Definition (
n
-connectedness).
X
is
n
-connected if for any
k ≤ n
, any continuous
f : S
k
→ X extends to a continuous F : D
k+1
→ X such that F |
S
k
= f.
For any point
p ∈ R
n
,
R
n
\{p}
is
m
-connected iff
m ≤ n −
2. So
R
n
\{p} 6'
R
m
\ {q} unless n = m. So R
n
6' R
m
.
Unfortunately, we have not yet proven that
R
n
6' R
m
. We casually stated
that
R
n
\ {p}
is
m
-connected iff
m ≤ n −
2. However, while this intuitively
makes sense, it is in fact very difficult to prove. To actually prove this, we will
need tools from algebraic topology.