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Unformatted text preview: Math 202A Fall 2005 David Penneys Problems by Sarason. (11) Let B be a Baire space. Let U B be an open subset. Let M U be a meager set in U . Since M is cleary meager in B , B \ M is residual in B and therefore dense in B since B is Baire. Let V U be an open subset in the relative topology for U , i.e., V = U W for some open W B . Since V is open in B as the intersection of two open sets in B , ( B \ M ) V is nonempty, and since V U , ( U \M ) V is nonempty. Thus U \M is dense, and U is a Baire space. (12) Let X be a T 1 Baire space with no isolated points. Since X is T 1 , all singletons are closed in X , and since X has no isolated points, all singletons are not open in X . Case 1 : X is counbtable. Let x X . { x } is meager, so X \{ x } is residual. But X \{ x } is the countable union of singletons, which are nowhere dense sets, so X \{ x } is meager and thus { x } is residual. But { x } is not dense since X \ { x } is open, and ( X \ { x } ) { x } is empty. Thus X is not a Baire space, a contradiction. Case 2 : X is uncountable. Suppose R X is residual. Suppose R is countable and has one point. Then X \ R is open, and R is not dense since ( X \ R ) R is empty. Thus X is not a Baire space, a contradiction. Suppose R is countable and has more than one point. Let x R . Then R \{ x } is meager as the countable union of singletons, which are nowhere dense. Thus ( X \ R ) ( R \ { x } ) is meager as the union of two meager sets (which is still the countable union of nowhere dense sets). So { x } = X \ (( X \ R ) ( R \ { x } )) is residual, but { x } is not dense as before, a contradiction. Thus, any residual subset must be uncountable. (13) Let X be a Baire space, f : X Y a continuous, open surjection. Let R Y be residual. We need to show R is dense in Y . We have that Y \ R is meager, so Y \ R = S n =1 C n where C n is nowhere dense n N . First, we show f 1 ( S n =1 C n ) is meager in X . To show this, it suffices to show that f 1 ( C n ) is meager since f 1 ( S n =1 C n ) = S n =1 f 1 ( C n ) and a countable union of meager sets is meager as the countable union of a countable union of nowhere dense sets, which is still a countable union of nowhere dense sets. Let U f 1 ( C n ) for some n N be an open subset. Then f ( U ) f ( f 1 ( C n )) f ( f 1 ( C n )) = C n where the first inclusion follows from the continuous image of a subset, the second inclusion follows from the defintion of continuity (a problem from the last homewrok set), and the equality follows from the fact that f is a surjection. But since f is an open map, f ( U ) is open, so int ( C n ) cannot be empty, a contradiction. Thus, there does not exist an open U f 1 ( C n ), so int ( f 1 ( C n )) is empty n N , and thus f 1 ( C n ) is meager n N , and f 1 ( S n =1 C n ) is meager in X . Thus f 1 ( R ) is dense in X ....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
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