204Lecture72006

# 204Lecture72006 - Economics 204 Lecture 7Tuesday August 8...

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Economics 204 Lecture 7–Tuesday, August 8, 2006 Revised 8/8/06, Revisions indicated by ** ** 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. 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. 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 1

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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. 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 X is connected, then f ( C ) is connected. Proof: We prove the contrapositive: if f ( C ) is not connected, then C is not connected. Suppose f ( C ) is not connected. Find P = = Q, f ( C ) = P Q, ¯ P Q = P ¯ Q = Let A = f 1 ( P ) C, B = f 1 ( Q ) C Then A B = f 1 ( P ) C f 1 ( Q ) C = f 1 ( P ) f 1 ( Q ) C = f 1 ( P Q ) C 2

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= f 1 ( f ( C )) C = C A = f 1 ( P ) C = B = f 1 ( Q ) C = A = f 1 ( P ) C f 1 ( P ) f 1 ¯ P which is closed, so ¯ A f 1 ¯ P B = f 1 ( Q ) C f 1 ( Q ) f 1 ¯ Q which is closed, so ¯ B f 1 ¯ Q ¯ A B f 1 ¯ P f 1 ( Q ) = f 1 ¯ P Q = f 1 ( ) = A ¯ B f 1 ( P ) f
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