3Connectivity

IB Metric and Topological Spaces

3.1 Connectivity

We will first look at normal connectivity.

Definition (Connected space). A topological space

X

is disconnected if

X

can

be written as

A ∪ B

, where

A

and

B

are disjoint, non-empty open subsets of

X

.

We say A and B disconnect X.

A space is connected if it is not disconnected.

Note that being connected is a property of a space, not a subset. When we

say “

A

is a connected subset of

X

”, it means

A

is connected with the subspace

topology inherited from X.

Being (dis)connected is a topological property, i.e. if

X

is (dis)connected,

and

X ' Y

, then

Y

is (dis)connected. To show this, let

f

:

X → Y

be the

homeomorphism. By definition,

A

is open in

X

iff

f

(

A

) is open in

Y

. So

A

and

B disconnect X iff f (A) and f(B) disconnect Y .

Example.

– If X has the coarse topology, it is connected.

– If X has the discrete topology and at least 2 elements, it is disconnected.

–

Let

X ⊆ R

. If there is some

α ∈ R \ X

such that there is some

a, b ∈ X

with

a < α < b

, then

X

is disconnected. In particular,

X ∩

(

−∞, α

) and

X ∩ (α, ∞) disconnect X.

For example, (0, 1) ∪ (1, 2) is disconnected (α = 1).

We can also characterize connectivity in terms of continuous functions:

Proposition.

X

is disconnected iff there exists a continuous surjective

f

:

X →

{0, 1} with the discrete topology.

Alternatively,

X

is connected iff any continuous map

f

:

X → {

0

,

1

}

is

constant.

Proof. (⇒) If A and B disconnect X, define

f(x) =

(

0 x ∈ A

1 x ∈ B

Then

f

−1

(

∅

) =

∅

,

f

−1

(

{

0

,

1

}

) =

X

,

f

−1

(

{

0

}

) =

A

and

f

−1

(

{

1

}

) =

B

are all

open. So f is continuous. Also, since A, B are non-empty, f is surjective.

(

⇐

) Given

f

:

X 7→ {

0

,

1

}

surjective and continuous, define

A

=

f

−1

(

{

0

}

),

B = f

−1

({1}). Then A and B disconnect X.

Theorem. [0, 1] is connected.

Note that

Q ∩

[0

,

1] is disconnected, since we can pick our favorite irrational

number

a

, and then

{x

:

x < a}

and

{x

:

x > a}

disconnect the interval. So we

better use something special about [0, 1].

The key property of R is that every non-empty A ⊆ [0, 1] has a supremum.

Proof.

Suppose

A

and

B

disconnect [0

,

1]. wlog, assume 1

∈ B

. Since

A

is

non-empty, α = sup A exists. Then either

– α ∈ A

. Then

α <

1, since 1

∈ B

. Since

A

is open,

∃ε >

0 with

B

ε

(

α

)

⊆ A

.

So α +

ε

2

∈ A, contradicting supremality of α; or

– α 6∈ A

. Then

α ∈ B

. Since

B

is open,

∃ε >

0 such that

B

ε

(

α

)

⊆ B

. Then

a ≤ α −ε

for all

a ∈ A

. This contradicts

α

being the least upper bound of

A.

Either option gives a contradiction. So

A

and

B

cannot exist and [0

,

1] is

connected.

To conclude the section, we will prove the intermediate value property. The

key proposition needed is the following.

Proposition. If

f

:

X → Y

is continuous and

X

is connected, then

im f

is also

connected.

Proof.

Suppose

A

and

B

disconnect

im f

. We will show that

f

−1

(

A

) and

f

−1

(

B

)

disconnect X.

Since

A, B ⊆ im f

are open, we know that

A

=

im f ∩A

0

and

B

=

im f ∩B

0

for some

A

0

, B

0

open in

Y

. Then

f

−1

(

A

) =

f

−1

(

A

0

) and

f

−1

(

B

) =

f

−1

(

B

0

) are

open in X.

Since

A, B

are non-empty,

f

−1

(

A

) and

f

−1

(

B

) are non-empty. Also,

f

−1

(

A

)

∩

f

−1

(

B

) =

f

−1

(

A ∩ B

) =

f

−1

(

∅

) =

∅

. Finally,

A ∪ B

=

im f

. So

f

−1

(

A

)

∪

f

−1

(B) = f

−1

(A ∪ B) = X.

So

f

−1

(

A

) and

f

−1

(

B

) disconnect

X

, contradicting our hypothesis. So

im f

is connected.

Alternatively, if

im f

is not connected, let

g

:

im f → {

0

,

1

}

be continuous

surjective. Then g ◦ f : X → {0, 1} is continuous surjective. Contradiction.

Theorem (Intermediate value theorem). Suppose

f

:

X → R

is continuous

and

X

is connected. If

∃x

0

, x

1

such that

f

(

x

0

)

<

0

< f

(

x

1

), then

∃x ∈ X

with

f(x) = 0.

Proof.

Suppose no such

x

exists. Then 0

6∈ im f

while 0

> f

(

x

0

)

∈ im f

,

0

< f

(

x

1

)

∈ im f

. Then

im f

is disconnected (from our previous example),

contradicting X being connected.

Alternatively, if

f

(

x

)

6

= 0 for all

x

, then

f(x)

|f(x)|

is a continuous surjection from

X to {−1, +1}, which is a contradiction.

Corollary. If

f

: [0

,

1]

→ R

is continuous with

f

(0)

<

0

< f

(1), then

∃x ∈

[0

,

1]

with f(x) = 0.

Is the converse of the intermediate value theorem true? If

X

is disconnected,

can we find a function g that doesn’t satisfy the intermediate value property?

The answer is yes. Since

X

is disconnected, let

f

:

X → {

0

,

1

}

be continu-

ous. Then let

g

(

x

) =

f

(

x

)

−

1

2

. Then

g

is continuous but doesn’t satisfy the

intermediate value property.