CONNECTED COMPONENTS
Recall the definition of
connectedness
(2.45 in [1]).
Definition 1.
Let
(
X, d
)
be a metric space. A subset
E
⊆
X
is called
disconnected
(or
sepa-
rated
) if there exists nonempty
A, B
⊂
E
such that
E
=
A
∪
B
, and
A
∩
B
=
A
∩
B
=
∅
. A
subset is called
connected
if it is not disconnected.
Example 2.
In
R
with the Euclidean metric, consider the set
E
= [0
,
1]
∪
(2
,
3)
.
Since
(2
,
3) = [2
,
3]
, the strong condition
[0
,
1]
∩
(2
,
3) =
∅
holds, which implies that the two
subsets
A
= [0
,
1]
and
B
= (2
,
3)
form a separation of
E
. Hence
E
is not connected.
Example 3.
Again in
R
, consider the set
E
= [0
,
1)
∪
(1
,
2]
. This set is also disconnected:
setting
A
= [0
,
1)
and
B
= (1
,
2]
we have
A
= [0
,
1]
which does not intersect
(1
,
2] =
B
,
while
(1
,
2] = [1
,
2]
does not intersect
[0
,
1) =
A
. Thus,
E
is disconnected. In this case,
A
∩
B
is not empty (it contains the point
1
). We might say that
A
and
B
are
adjacent
. But
they are not connected. A person living in
A
cannot communicate with a person living in
B
without sending a message outside the universe
E
=
A
∪
B
.
Example 4.
Once more in
R
, take
E
= [0
,
2]
. We can express
E
as a union of two non-
intersecting pieces, for example
E
= [0
,
1)
∪
[1
,
2]
. (Indeed, any set with at least two points
can be divided into two non-empty pieces.) But in this case,
[0
,
1)
∩
[1
,
2]
is non-empty (it
consists of the single point
1
), and hence this does not constitute a separation of
E
. The
point of Definition 1 is that, to be disconnected, a set must consist of (at least) two pieces
that
do not touch
. In fact, the set
[0
,
2]
is
connected, as proved in Theorem 2.47 in [1]: the
connected subsets of
R
are precisely the intervals.
The concept of connectedness can be used to naturally divide any metric space into a
collection of pieces, called its
components
. To see how to do this, we need the following
lemma.
Lemma 5.
Let
G
be any collection of connected subsets of a metric space
(
X, d
)
. Suppose that
T
G
is not empty. Then
S
G
is connected.
Remark
6
.
The assumptions of Lemma 5 are stronger than needed to deduce the conclu-
sion. For example, all we really need to assume is that, for any two sets
U, V
∈
G
either
U
∩
V
or
U
∩
V
is non-empty (i.e. any two sets in
G
touch). It will then following that the
union of all sets in
G
is connected. The proof of this stronger theorem (i.e. with a weaker
hypothesis) is only slightly more complicated than the proof below; but we only need the
statement of Lemma 5 in what follows, so we’ll stick with what we need.
Proof.
To produce a contradiction, let us suppose that
S
G
is disconnected. That is, let
A, B
be two non-empty sets so that
S
G
=
A
∪
B
, and
A
∩
B
=
A
∩
B
=
∅
. Let
U
be
any set in
G
. Note that
U
∩
A
⊆
A
and
U
∩
B
⊆
B
(you should work out why this is
true). Hence, since
A
∩
B
=
∅
, it follows that
U
∩
A
∩
(
U
∩
B
)
⊆
A
∩
B
=
∅
, and similarly
(
U
∩
A
)
∩
U
∩
B
=
∅
. But
U
∈
G
so
U
⊆
S
G
=
A
∪
B
; this means that
(
U
∩
A
)
∪
(
U
∩
B
) =
U
.