6. A glimpse at metric spaces 23
Definition 6.3. Let (X,δ) be a metric space. A subset S of X is called connected
if the following holds: whenever S = A∪B, where A and B are open and A∩B = ∅,
then either A = or B = ∅.
In Chapter 6 we will use the following fairly simple result. In it and in the
subsequent comment we assume C is equipped with the usual metric.
Theorem 6.1. An open subset of C is connected if and only if each pair of
points z,w can be joined by a polygonal path whose sides are parallel to the
axes.
Proof. (Sketch.) The result holds (and is uninteresting) when is the empty set.
We therefore first assume is connected and nonempty. Choose z Ω. Consider
the set S of points that can be reached from z by such a polygonal path. Then
S is not empty, as z S. Also, S is open; if we can reach w, then we can also
reach points in a ball about w. On the other hand, the set T of points we cannot
reach from z is also open. By the definition of connectedness, is not the union of
disjoint open subsets unless one of them is empty. Thus, as S is nonempty, T must
be empty. Hence S = and the conclusion holds.
The proof of the converse statement is similar; we prove its contrapositive.
Assume is not connected and nonempty. Then = A B, where A and B are
open, A∩B = ∅, but neither A nor B is empty. One then checks that no polygonal
path in connects points in A to points in B. Hence the contrapositive of the
converse statement holds.
We pause to state the intuitive characterization of connected subsets of the real
line. A subset of R (in the usual metric) is connected if and only if it is an interval.
By convention, the word interval includes the entire real line, the empty set, and
semi-infinite intervals. We leave the proof as an exercise, with the following hints.
Given a bounded connected set S, let α be its greatest lower bound and let β be
its least upper bound. We claim that S must be the interval between α and β,
perhaps including one or both of these end points. To check this assertion, suppose
α x0 β. If x0 is not in S, then {x S : x x0} and {x S : x x0} are open
nonempty subsets of S violating the definition of connectedness. The same idea
works when S is unbounded below or above. The converse (an interval is connected)
proceeds by writing the interval as A∪B, where A and B are nonemptyclosed sets.
(Their complements are open.) Choosing points in each and successively bisecting
the interval between them creates two monotone sequences with a common limit,
which must then be in both sets. Hence A B is nonempty.
I
Exercise 1.36. Complete the proof that a subset of R is connected if and only
if it is an interval.
Finally we mention one more concept, distinct from connectedness, but with a
similar name. Roughly speaking, an open and connected subset S of C is called
simply connected if it has no holes. Intuitively speaking, S has a hole if there is a
closed curve in S which surrounds at least one point not in S. For example, the
complement of a point is open and connected, but it is not simply connected. The
set of z for which 1 |z| 2 is open and connected but not simply connected. In
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