Unformatted text preview: < f ( a ) . Proof. f ([ a,b ]) is an interval, since it is connected and since any connected set in R is an interval. Suppose f ( a ) < f ( b ). Since f ( a ) ∈ f ([ a,b ]) and f ( b ) ∈ f ([ a,b ]), it follows that [ f ( a ) ,f ( b )] ⊆ f ([ a,b ]). So, if y ∈ ( f ( a ) ,f ( b )) then y ∈ f ([ a,b ]) and so there is an x ∈ [ a,b ] such that f ( x ) = y . Since y 6 = f ( a ) and y 6 = f ( b ), x ∈ ( a,b ). The argument for f ( b ) < f ( a ) is similar. ± Example 1 . Suppose f ( a ) < 0 and f ( b ) > 0. Then there is an x ∈ ( a,b ) such that f ( x ) = 0. This fact can be used to show the existence of a competitive equilibrium if there are only 2 commodities. To handle the case of more than 2 commodities, more sophisticated machinery must be employed (in particular, the Brouwer ﬁxed point theorem). ² 1...
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This note was uploaded on 05/20/2010 for the course ECON 511 taught by Professor Johnnachbar during the Fall '08 term at Washington University in St. Louis.
 Fall '08
 JohnNachbar

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