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.