102 Chapter 2
which is a representation of A as the union of two disjoint nonempty closed
sets with respect to the subspace A. Since this is impossible, A is connected.
(5) Suppose that A is disconnected. Then we have AC G 1 U G 2 , where
G 1 and G 2 are disjoint open sets such that AnG 1 and AnG 2 are nonempty.
Let p be any point of A n G 1 and q any point of A n G 2 • By assumption
there is a connected subset C of A containing both points p and q. However,
C c A c G 1 U G 2 and the sets C n G 1 , C n G 2 are not empty since they
contain the points p and q, respectively. This is impossible because C is
connected. Therefore, A is connected.
It fol.lows from Theorem 2.8-3 that every point a of a set A in a met-
ric space belongs to a maximal connected subset that is the union of all
connected subsets of A containing the point a. This leads to the following
definition.
Definition 2.31 The largest connected subset containing a point a E A
is called the component of A determined by a. By Theorem 2.8-4, any two
components of A are either disjoint or identical. Hence there is a unique
decomposition of any set into components, and we may define a compo-
nent C of a set as a connected subset that is not contained in any larger
connected subset (i.e., without reference to any particular point a E C).
If A is a connected set, it has only one component, namely, A itself. At
the other extreme, it may happen that every component of A reduces to a
point. In this case A is said to be a totally disconnected set.
If ( S, d) is a metric space, the preceding discussion can be applied to the
set S, so we may consider S divided into components.
Theorem 2.9 The components of a metric space are closed sets.
Proof It follows at once from Theorem 2.8-1.
Theorem 2.10 The real line is connected, and the only connected subsets
of the real line are the intervals.
In the statement above, intervals of the form [a, a] = {a} are admissible.
Proof Suppose that the real line R is disconnected.. Then we may write
R = AU B, where A and B are nonempty disjoint closed sets. Choosea1 EA and b1 EB arbitrarily and let m be the midpoint of [a 1 , b 1 ]. Since
m E R we have either m E A or m E B. In the first case consider the
interval [m, b 1 ] and call it [a 2 , b 2 ]; in the second case consider the interval
[a1, m] and call it [a 2 , b2]. By continuing the process indefinitely we obtaina sequence {[an, bn]} of nested intervals with an E A, bn EB for all n. vVe
know that the sequences {an} and {bn} have a common limit L. Since A
and B are closed we have L E A and L E B, which is impossible since A
and B are disjoint. Therefore, R is connected.