10.2. CONNECTED SETS 259

10.2 Connected SetsThis has absolutely nothing to do with linear algebra but is here to provide convenientresults to be used later when linear algebra will occur as part of some topics in analysis.

Stated informally, connected sets are those which are in one piece. In order to definewhat is meant by this, I will first consider what it means for a set to not be in one piece.This is called separated. Connected sets are defined in terms of not being separated. Thisis why theorems about connected sets sometimes seem a little tricky.

Definition 10.2.1 A set, S in a metric space, is separated if there exist sets A,B such that

S = A∪B, A,B ̸= /0, and A∩B = B∩A = /0.

In this case, the sets A and B are said to separate S. A set is connected if it is not separated.Remember A denotes the closure of the set A.

Note that the concept of connected sets is defined in terms of what it is not. This makesit somewhat difficult to understand. One of the most important theorems about connectedsets is the following.

Theorem 10.2.2 Suppose U is a set of connected sets and that there exists a point p whichis in all of these connected sets. Then K ≡ ∪U is connected.

Proof: Suppose K = A∪B where Ā∩B = B̄∩A = /0,A ̸= /0,B ̸= /0. Let U ∈U . ThenU = (U ∩A)∪ (U ∩B) and this would separate U if both sets in the union are nonemptysince the limit points of U ∩B are contained in the limit points of B. It follows that everyset of U is contained in one of A or B. Suppose then that some U ⊆ A. Then all U ∈ Umust be contained in A because if one is contained in B, this would violate the assumptionthat they all have a point p in common. Thus K is connected after all because this requiresB = /0. Alternatively, p is in one of these sets. Say p ∈ A. Then by the above argumentevery U must be in A because if not, the above would be a separation of U . Thus B = /0. ■

The intersection of connected sets is not necessarily connected as is shown by the fol-lowing picture.

U

V

Theorem 10.2.3 Let f : X → Y be continuous where Y is a metric space and X is con-nected. Then f (X) is also connected.

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 anopen set, U containing f (z) such that U∩A= /0. But then, the continuity of f and Theorem