204Lecture72008Web

# 204Lecture72008Web - Economics 204 Lecture 7Tuesday, August...

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Economics 204 Lecture 7–Tuesday, August 5, 2008 Note : In this set of lecture notes, ¯ A refers to the closure of A . Section 2.9, Connected Sets Defnition 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 } 6 = [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) Ase t 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 contra- positive: if S is not an interval, it is not connected. If S is not an interval, ±nd x, y S, x < z < y, z 6∈ S 1

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Let A = S ( −∞ ,z ) ,B = S ( z, ) Then ¯ A B ( −∞ ) ( )=( −∞ ] ( )= A ¯ B ( −∞ ) ( −∞ ) [ A B =( S ( −∞ )) ( S ( )) = S \{ z } = S x A, so A 6 = y B, so B 6 = 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 continuous. 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 6 = ∅ 6 = 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 ² ± f 1 ( Q ) C ² = ± f 1 ( P ) f 1 ( Q ) ² C = f 1 ( P Q ) C 2

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= f 1 ( f ( C )) C = C A = f 1 ( P ) C 6 = B = f 1 ( Q ) C 6 = A = f 1 ( P ) C f 1 ( P ) f 1 ± ¯ P ² which is closed, so ¯ A f 1 ± ¯ P ² B = f 1 ( Q ) C f 1 ( Q ) f 1 ± ¯ Q ² which is closed, so ¯ B f 1 ± ¯ Q ² ¯ A B f 1 ± ¯ P ² f 1 ( Q ) = f 1 ± ¯ P Q ² = f 1 ( ) = A ¯ B f 1 ( P ) f 1 ± ¯ Q ² = f 1 ± P ¯ Q ² = f 1 ( ) = so C is not connected.
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## 204Lecture72008Web - Economics 204 Lecture 7Tuesday, August...

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