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 Ω of C is connected if and only if each pair of

points z,w ∈ Ω can be joined by a polygonal path whose sides are parallel to the

axes.

Proof. (Sketch.) The result holds (and is uninteresting) when Ω is the empty set.

We therefore ﬁrst assume Ω 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, Ω 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 = Ω and the conclusion holds.

The proof of the converse statement is similar; we prove its contrapositive.

Assume Ω is not connected and nonempty. Then Ω = 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 Ω 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