3Connectivity
IB Metric and Topological Spaces
3.3 Components
If a space is disconnected, we could divide the space into different components,
each of which is (maximally) connected. This is what we will investigate in this
section. Of course, we would have a different notion of “components” for each
type of connectivity. We will first start with path connectivity.
3.3.1 Path components
Defining path components (i.e. components with respect to path connectedness)
is relatively easy. To do so, we need the following lemma.
Lemma. Define
x ∼ y
if there is a path from
x
to
y
in
X
. Then
∼
is an
equivalence relation.
Proof.
(i)
For any
x ∈ X
, let
γ
x
: [0
,
1]
→ X
be
γ
(
t
) =
x
, the constant path. Then
this is a path from x to x. So x ∼ x.
(ii)
If
γ
: [0
,
1]
→ X
is a path from
x
to
y
, then
¯γ
: [0
,
1]
→ X
by
t 7→ γ
(1
− t
)
is a path from y to x. So x ∼ y ⇒ y ∼ x.
(iii)
If
γ
1
is a path from
x
to
y
and
γ
2
is a path from
y
to
z
, then
γ
2
∗γ
1
defined
by
t 7→
(
γ
1
(2t) t ∈ [0, 1/2]
γ
2
(2t − 1) t ∈ [1/2, 1]
is a path from x to z. So x ∼ y, y ∼ z ⇒ x ∼ z.
With this lemma, we can immediately define the path components:
Definition (Path components). Equivalence classes of the relation “
x ∼ y
if
there is a path from x to y” are path components of X.
3.3.2 Connected components
Defining connected components (i.e. components with respect to regular connec-
tivity) is slightly more involved. The key proposition we need is as follows:
Proposition. Suppose
Y
α
⊆ X
is connected for all
α ∈ T
and that
T
α∈T
Y
α
6
=
∅
.
Then Y =
S
α∈T
Y
α
is connected.
Proof.
Suppose the contrary that
A
and
B
disconnect
Y
. Then
A
and
B
are
open in
Y
. So
A
=
Y ∩ A
0
and
B
=
Y ∩ B
0
, where
A
0
, B
0
are open in
X
. For
any fixed α, let
A
α
= Y
α
∩ A = Y
α
∩ A
0
, B
α
= Y
α
∩ B = Y
α
∩ B
0
.
Then they are open in Y
α
. Since Y = A ∪ B, we have
Y
α
= Y ∩Y
α
= (A ∪ B) ∩ Y
α
= A
α
∪ B
α
.
Since A ∩ B = ∅, we have
A
α
∩ B
α
= Y
α
∩ (A ∩ B) = ∅.
So
A
α
, B
α
are disjoint. So
Y
α
is connected but is the disjoint union of open
subsets A
α
, B
α
.
By definition of connectivity, this can only happen if A
α
= ∅ or B
α
= ∅.
However, by assumption,
\
α∈T
Y
α
6
=
∅
. So pick
y ∈
\
α∈T
Y
α
. Since
y ∈ Y
,
either
y ∈ A
or
y ∈ B
. wlog, assume
y ∈ A
. Then
y ∈ Y
α
for all
α
implies that
y ∈ A
α
for all
α
. So
A
α
is non-empty for all
α
. So
B
α
is empty for all
α
. So
B = ∅.
So A and B did not disconnect Y after all. Contradiction.
Using this lemma, we can define connected components:
Definition (Connected component). If x ∈ X, define
C(x) = {A ⊆ X : x ∈ A and A is connected}.
Then C(x) =
[
A∈C(x)
A is the connected component of x.
C
(
x
) is the largest connected subset of
X
containing
x
. To show this, we
first note that
{x} ∈ C
(
x
). So
x ∈ C
(
x
). To show that it is connected, just note
that
x ∈
T
A∈C(x)
A
. So
T
A∈C(x)
A
is non-empty. By our previous proposition,
this implies that C(x) is connected.
Lemma. If y ∈ C(x), then C(y) = C(x).
Proof.
Since
y ∈ C
(
x
) and
C
(
x
) is connected,
C
(
x
)
⊆ C
(
y
). So
x ∈ C
(
y
). Then
C(y) ⊆ C(x). So C(x) = C(y).
It follows that
x ∼ y
if
x ∈ C
(
y
) is an equivalence relation and the connected
components of X are the equivalence classes.
Example.
–
Let
X
= (
−∞,
0)
∪
(0
, ∞
)
⊆ R
. Then the connected components are
(−∞, 0) and (0, ∞), which are also the path components.
–
Let
X
=
Q ⊆ R
. Then
C
(
x
) =
{x}
for all
x ∈ X
. In this case, we say
X
is totally disconnected.
Note that
C
(
x
) and
X \C
(
x
) need not disconnect
X
, even though it is the
case in our first example. For this, we must need
C
(
x
) and
X \C
(
x
) to be open
as well. For example, in Example 2, C(x) = {x} is not open.
It is also important to note that path components need not be equal to the
connected components, as illustrated by the following example. However, since
path connected spaces are connected, the path component containing
x
must be
a subset of C(x).
Example. Let Y = {(0, y) : y ∈ R} ⊆ R
2
be the y axis.
Let Z = {(x,
1
x
sin
1
x
) : x ∈ (0, ∞)}.
x
y
Y
Z
Let
X
=
Y ∪Z ⊆ R
2
. We claim that
Y
and
Z
are the path components of
X
.
Since
Y
and
Z
are individually path connected, it suffices to show that there is
no continuous γ : [0, 1] → X with γ(0) = (0, 0), γ(1) = (1, sin 1).
Suppose
γ
existed. Then the function
π
2
◦ γ
: [0
,
1]
→ R
projecting the
path to the
y
direction is continuous. So it is bounded. Let
M
be such that
π
2
◦ γ
(
t
)
≤ M
for all
t ∈
[0
,
1]. Let
W
=
X ∩
(
R ×
(
−∞, M
]) be the part of
X
that lies below y = M. Then im γ ⊆ W .
However,
W
is disconnected: pick
t
0
with
1
t
0
sin
1
t
0
> M
. Then
W ∩
((
−∞, t
0
)
× R
) and
W ∩
((
t
0
, ∞
)
× R
) disconnect
W
. This is a contradiction,
since γ is continuous and [0, 1] is connected.
We also claim that
X
is connected: suppose
A
and
B
disconnect
X
. Then
since
Y
and
Z
are connected, either
Y ⊆ A
or
Y ⊆ B
;
Z ⊆ A
or
Z ⊆ B
. If both
Y ⊆ A, Z ⊆ A, then B = ∅, which is not possible.
So wlog assume
A
=
Y
,
B
=
Z
. This is also impossible, since
Y
is not open
in
X
as it is not a union of balls (any open ball containing a point in
Y
will also
contain a point in Z). Hence X must be connected.
Finally, recall that we showed that path-connected subsets are connected.
While the converse is not true in general, there are special cases where it is true.
Proposition. If U ⊆ R
n
is open and connected, then it is path-connected.
Proof. Let A be a path component of U. We first show that A is open.
Let
a ∈ A
. Since
U
is open,
∃ε >
0 such that
B
ε
(
a
)
⊆ U
. We know that
B
ε
(
a
)
' Int
(
D
n
) is path-connected (e.g. use line segments connecting the points).
Since
A
is a path component and
a ∈ A
, we must have
B
ε
(
a
)
⊆ A
. So
A
is an
open subset of U .
Now suppose b ∈ U \ A. Then since U is open, ∃ε > 0 such that B
ε
(b) ⊆ U.
Since
B
ε
(
b
) is path-connected, so if
B
ε
(
b
)
∩ A 6
=
∅
, then
B
ε
(
b
)
⊆ A
. But this
implies
b ∈ A
, which is a contradiction. So
B
ε
(
b
)
∩ A
=
∅
. So
B
ε
(
b
)
⊆ U \ A
.
Then U \ A is open.
So
A, U \ A
are disjoint open subsets of
U
. Since
U
is connected, we must
have U \ A empty (since A is not). So U = A is path-connected.