Multivariable Advanced Calculus

5.3. CONNECTED SETS 91The intersection of connected sets is not necessarily connected as is shown by thefollowing picture.UVTheorem 5.3.5 Let f : X → Y be continuous where Y is a normed vector spaceand X is connected. Then f (X) is also 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) andf −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 open set, U containing f (z) such that U ∩A = ∅. But then, the continuityof f and Theorem 5.0.2 implies that f −1 (U) is an open set containing z such thatf −1 (U) ∩ f −1 (A) = ∅. Therefore, f −1 (B) contains no limit points of f −1 (A) . Similarreasoning implies f −1 (A) contains no limit points of f −1 (B). It follows that X isseparated by f −1 (A) and f −1 (B) , contradicting the assumption that X was 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 5.3.6 Let S be a set and let p ∈ S. Denote by C pthe union ofall connected subsets of S which contain p. This is called the connected componentdetermined by p.Theorem 5.3.7 Let C p be a connected component of a set S in a normed vectorspace. Then C p is a connected set and if C p ∩ C q ≠ ∅, then C p = C q .Proof: Let C denote the connected subsets of S which contain p. If C p = A ∪ BwhereA ∩ B = B ∩ A = ∅,then p is in one of A or B. Suppose without loss of generality p ∈ A. Then every setof C must also be contained in A since otherwise, as in Theorem 5.3.4, the set wouldbe separated. But this implies B is empty. Therefore, C p is connected. From this, andTheorem 5.3.4, the second assertion of the theorem is proved.This shows the connected components of a set are equivalence classes and partitionthe set.A set, I is an interval in R if and only if whenever x, y ∈ I then (x, y) ⊆ I. Thefollowing theorem is about the connected sets in R.Theorem 5.3.8 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 nothingto 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