This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem Set 3 Economics 204  August 2006 1. Connectedness (a) Suppose we define two sets as being separated if neither contains a limit point of the other and that a set in a metric space is connected if it cannot be written as the union of two disjoint nonempty separated sets. Prove that this definition is equivalent to the one given in lecture. To avoid confusion we will refer to the definition of separated and connected from the textbook as textseparated and textconnected and the definitions intro duced here will be referred to as psseparated and psconnected . It is tempting to try to show that psseparated equivalent to the textseparated. If we could show this, because all textseparated are disjoint we would know that the two definitions of connected are equivalent. However, the two definitions of separated are not equivalent. An isolated point of a set is not a limit point. Thus, if A and B have an isolated point in common (e.g. A = (0 , 1) { 2 } and B = { 2 } (3 , 4)) they can be psseparated but not disjoint (and therefore not textseparated). The definition of psconnected excludes this possiblity by specifying that the separated sets must be disjoint. Now proceed with the proof that a set is psconnected if and only if it is textconnected by showing that two disjoint sets A and B are psseparated if and only if they are textseparated. Proof: ( ) We will prove the contrapositive. Suppose that A and B are not textseparated. Without loss of generality, suppose that that there exists an x A cl B . Because x cl B , for all > 0 there exists b B ( x ) B . This means that either x is a limit point or an isolated point of B . If it is a limit point then we are done, because this means that A contains a limit point of B and therefore A and B are not psseparated. If x an isolated point of B , it cannot be an isolated point of A , because the sets are disjoint. This means that x must be a limit point of A , so B contains a limit point of A and the sets cannot be psseparated. ( ) Again we prove the contrapositive. Suppose that the sets are not psseparated. Without loss of generality, suppose that there exists a b L A which is a limit point of B . By definition, for all > 0, B ( b L ) contains some element of B , so it also must be in the closure of B . Therefore A cl B 6 = so A and B are not textseparated. (b) Suppose that the sets C and D are two nonempty separated subsets of X whose union is X and Y is a connected subset of X . Prove that Y lies entirely within C or D . Proof: Being separated, C and cl D must be disjoint as must be cl C and D . The sets cl C Y and D Y must also be disjoint and their union is Y so they are a separation of Y and the same goes for C Y and cl D Y . For each of these pairs, both sets being nonempty contradicts the fact that Y is connected. Therefore 1 one of them is empty. Therefore Y lies entirely within C or D ....
View
Full
Document
 Summer '08
 Staff
 Economics

Click to edit the document details