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.