6.5. CONNECTED SETS 109

Theorem 6.5.5 Let f : X → R be continuous where X is connected. Then f (X) isalso connected.

Proof: To do this you show f (X) is not separated. Suppose to the contrary thatf (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 existsan open ball U of radius ε for some ε > 0 containing f (z) such that U ∩A = /0. But then,the continuity of f and the definition of continuity imply that there exists δ > 0 such thatf (B(z,δ )) ⊆U . Therefore z is not a limit point of f−1 (A) . Since z was arbitrary, it fol-lows that f−1 (B) contains no limit 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 X was connected.

On R the connected sets are pretty easy to describe. A set, I is an interval in R if andonly if whenever x,y ∈ I then (x,y)⊆ I. The following theorem is about the connected setsin R.

Theorem 6.5.6 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. Ifx ∈ (p,q)\C, let C∩ (−∞,x)≡ A, and C∩ (x,∞)≡ B. Then C = A∪B and the sets A andB separate C contrary to the assumption that C is connected.

Conversely, let I be an interval. Suppose I is separated by A and B. Pick x∈A and y∈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 bound for S. Therefore, l ∈ B which impliesl /∈ A after all, a contradiction. It follows I must be connected.

Another useful idea is that of connected components. An arbitrary set can be writtenas a union of maximal connected sets called connected components. This is the concept ofthe next definition.

Definition 6.5.7 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 6.5.8 Let Cp be a connected component of a set S. Then Cp is a connectedset and if Cp∩Cq ̸= /0, then Cp =Cq.

Proof: Let C denote the connected subsets of S which contain p. If Cp = A∪B whereA∩B = B∩A = /0, then p is in one of A or B. Suppose without loss of generality p ∈ A.Then every set of C must also be contained in A since otherwise, as in Theorem 6.5.4, theset would be separated. But this implies B is empty. Therefore, Cp is connected. From this,and Theorem 6.5.4, the second assertion of the theorem is proved.

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

Probably the most useful application of this is to the case where you have an open setand consider its connected components.