2.10. CONNECTED SETS 71

Proof: To do this you show f(X) is not separated. Suppose to the contrary that f(X) =A∪B where A and B separate f(X) . Then consider the sets f−1 (A) and f−1 (B) . If z ∈f−1 (B) , then f(z) ∈ B and so f(z) is not a limit point of A. Therefore, there exists an openset, U containing f(z) such that U ∩A = /0. But then, the continuity of f implies that f−1 (U)is an open set containing z such that f−1 (U)∩ f−1 (A) = /0. Therefore, f−1 (B) contains nolimit points of f−1 (A) . Similar reasoning implies f−1 (A) contains no limit points of f−1 (B).It follows that X is separated by f−1 (A) and f−1 (B) , contradicting the assumption that Xwas connected. ■

An arbitrary set can be written as a union of maximal connected sets called connectedcomponents. This is the concept of the next definition.

Definition 2.10.4 Let S be a set and let p ∈ S. Denote by Cp the union of all con-nected subsets of S which contain p. This is called the connected component determined byp.

Theorem 2.10.5 Let Cp be a connected component of a set S . Then Cp is a con-nected set and if Cp∩Cq ̸= /0, then Cp =Cq.

Proof: Let C denote the connected subsets of S which contain p. By Theorem 2.10.2,∪C = Cp is connected. If x ∈ Cp ∩Cq, then from Theorem 2.10.2, Cp ⊇ Cp ∪Cq and soCp ⊇Cq. The inclusion goes the other way by the same reason. ■

This shows the connected components of a set are equivalence classes and partition theset.

A set I is an interval in R if and only if whenever x,y ∈ I then [x,y]⊆ I. The followingtheorem is about the connected sets in R.

Theorem 2.10.6 A set C in R is connected if and only if C is an interval.

Proof: Let C be connected. If C consists of a single point p, there is nothing to prove.The interval is just [p, p] . Suppose p < q and p,q ∈C. You need to show (p,q)⊆C. If

x ∈ (p,q)\C

let C∩ (−∞,x) ≡ A, and C∩ (x,∞) ≡ B. Then C = A∪B and the sets A and B separate Ccontrary to the assumption that C is connected.

Conversely, let I be an interval. Suppose I is separated by A and B. Pick x ∈ A andy ∈ B. Suppose without loss of generality that x < y. Now define the set,

S≡ {t ∈ [x,y] : [x, t]⊆ A}

and let l be the least upper bound of S. Then l ∈ A so l /∈ B which implies l ∈ A. But if l /∈ B,then for some δ > 0, (l, l +δ )∩B = /0. contradicting the definition of l as an upper boundfor S. Therefore, l ∈ B which implies l /∈ A after all, a contradiction. It follows I must beconnected. ■

This yields a generalization of the intermediate value theorem from one variable calcu-lus.

Corollary 2.10.7 Let E be a connected set in Rp and suppose f : E → R and thaty ∈ ( f (e1) , f (e2)) where ei ∈ E. Then there exists e ∈ E such that f (e) = y.