204ps32006sols - Problem Set 3 Economics 204 - August 2006...

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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 non-empty 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 text-separated and text-connected and the definitions intro- duced here will be referred to as ps-separated and ps-connected . It is tempting to try to show that ps-separated equivalent to the text-separated. If we could show this, because all text-separated 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 ps-separated but not disjoint (and therefore not text-separated). The definition of ps-connected excludes this possiblity by specifying that the separated sets must be disjoint. Now proceed with the proof that a set is ps-connected if and only if it is text-connected by showing that two disjoint sets A and B are ps-separated if and only if they are text-separated. Proof: ( ) We will prove the contrapositive. Suppose that A and B are not text-separated. 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 ps-separated. 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 ps-separated. ( ) Again we prove the contrapositive. Suppose that the sets are not ps-separated. 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 text-separated. (b) Suppose that the sets C and D are two non-empty 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 non-empty contradicts the fact that Y is connected. Therefore 1 one of them is empty. Therefore Y lies entirely within C or D ....
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204ps32006sols - Problem Set 3 Economics 204 - August 2006...

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