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# Ill show first that f is continuous let x u we

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I’ll show first that f is continuous. Let x 0 U . We consider three cases. Case 1. x 0 is an interior point of U ; that is, it is not an endpoint of U . Then f ( x 0 ) cannot be an endpoint of the interval V . In fact, U contains points less than and larger than x 0 ; these points must map onto points respectively less than and larger than f ( x 0 ) since f is strictly increasing. Let ² > 0 be given. Possibly one (or both) of f ( x 0 ) + ² or f ( x 0 ) - ² is not in V , but there is η , 0 < η ² such that [ f ( x 0 ) - η, f ( x 0 ) + η ] V . Since f is onto, there exist points x 1 , x 2 U such that f ( x 1 ) = f ( x 0 ) - η , f ( x 2 ) = f ( x 0 ) + η . Because f is strictly increasing, we must have x 1 < x 0 < x 2 and f ( x ) ( f ( x 0 ) - η, f ( x 0 ) + η ) if x 1 < x < x 2 . Let δ = min( x 0 - x 1 , x 2 - x 0 ). Then δ > 0 and if | x - x 0 | < δ , then x 1 < x < x 2 , thus f ( x ) ( f ( x 0 ) - η, f ( x 0 ) + η ) and | f ( x ) - f ( x 0 ) | < η ². This takes care of continuity in Case 1.

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Case 2. x 0 is the right end-point of U . Because f is strictly increasing, f ( x 0 ) is easily seen to be the right end-point of V . The argument is then essentially as before. Let ² > 0 be given. There is then η > 0, η ² such that f ( x 0 ) - η V , hence [ f ( x 0 ) - η, f ( x 0 )] V . There is then x 1 U with f ( x 1 ) = f ( x 0 ) - η and the whole interval [ x 1 , x 0 ] maps into (actually onto) [ f ( x 0 ) - η, f ( x 0 )]. If δ = x 0 - x 1 then δ > 0 and x U , | x - x 0 | δ implies | f ( x ) - f ( x 0 ) | < η ² . Case 3. x 0 is the left end-point of U . Because f is strictly increasing, f ( x 0 ) is easily seen to be the left end-point of V and the argument is, mutatis mutandis, the same as in Case 2. Having established the continuity of f , the one of f - 1 follows. If y 1 , y 2 V and y 1 < y 2 , let x 1 , x 2 U with f ( x 1 ) = y 1 , f ( x 2 ) = y 2 . If x 1 x 2 , then y 1 y 2 , thus we must have x 1 < x 2 . This proves that f - 1 : V U is strictly increasing; that it is onto is a consequence of the fact that U is the domain of f . Thus f - 1 satisfies the same assumptions as f hence is continuous.
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• Fall '09
• Schonbek
• Topology, Metric space, Compact space, Closed set, ∂S

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