204Lecture72009

# 204Lecture72009 - Economics 204 Lecture 7Tuesday August 4...

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Economics 204 Lecture 7–Tuesday, August 4, 2009 Revised 8/5/09, Revisions indicated by ** and Sticky Notes Note : In this set of lecture notes, ¯ A refers to the closure of A . Section 2.9, Connected Sets Definition 1 Two sets A, B in a metric space are separated if ¯ A B = A ¯ B = A set in a metric space is connected if it cannot be written as the union of two nonempty separated sets. ** Remark: In other texts, you will see the following equivalent definition: A set Y in a metric space X is connected if there do not exist open sets A and B such that A B = , Y A B and A Y = and B Y = . Example: [0 , 1) and [1 , 2] are disjoint but not separated: [0 , 1) [1 , 2] = [0 , 1] [1 , 2] = { 1 } = [0 , 1) and (1 , 2] are separated: [0 , 1) (1 , 2] = [0 , 1] (1 , 2] = [0 , 1) (1 , 2] = [0 , 1) [1 , 2] = Note that d ([0 , 1) , (1 , 2]) = 0 even though the sets are separated. Note that separation does not require that ¯ A ¯ B = . [0 , 1) (1 , 2] is not connected. Theorem 2 (9.2) A set S of real numbers is connected if and only if it is an interval, i.e. given x, y S and z ( x, y ) , then z S . 1

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Proof: First, we show that S connected implies that S is an interval. We do this by proving the contrapositive: if S is not an interval, it is not connected. If S is not an interval, find x, y S, x < z < y, z S Let A = S ( −∞ , z ) , B = S ( z, ) Then ¯ A B ( −∞ , z ) ( z, ) = ( −∞ , z ] ( z, ) = A ¯ B ( −∞ , z ) ( z, ) = ( −∞ , z ) [ z, ) = A B = ( S ( −∞ , z )) ( S ( z, )) = S \ { z } = S x A, so A = y B, so B = so S is not connected. We have shown that if S is not an interval, then S is not connected; therefore, if S is connected, then S is an interval. Now, we need to show that if S is an interval, it is connected. This is much like the proof of the Intermediate Value Theorem. See de la Fuente for the details. Theorem 3 (9.3) Let X be a metric space, f : X Y con- tinuous. If C is a connected subset of X , then f ( C ) is con- nected. Proof: This is problem 5(b) on Problem Set 3. The idea is in the diagram. Prove the contrapositive: if f ( C ) is not connected, then C is not connected. 2
Corollary 4 (Intermediate Value Theorem) If f : [ a, b ] R is continuous, and f ( a ) < d < f ( b ) , then there exists c ( a, b ) such that f ( c ) = d . Proof: This is our third, and slickest, proof of the Intermediate Value Theorem. It is short because a substantial part of the proof was incorporated into the proof that C R is connected if and only if C is an interval, and the proof that if C is connected, then f ( C ) is connected. Here’s the proof: [ a, b ] is an interval, so [ a, b ] is connected, so f ([ a, b ]) is connected, so f ([ a, b ]) is an interval.

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