202A - Math 202A Fall 2005 David Penneys Problems by...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 26

202A - Math 202A Fall 2005 David Penneys Problems by...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online