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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 6 Rik Sengupta rsengupt@princeton.edu March 30, 2010 1. Munkres, p. 199, problem 3 Show that every order topology is regular. Solution. We first prove that every order topology is Hausdorff. So let ( X, ) be a simply ordered set. Let X be equipped with the order topology induced by the simple order. Fur thermore, let a and b be two distinct points in X , and suppose without loss of generality that a < b . Let A = { x X : a < x < b } , i.e. the set of elements between a and b . So now, if A is empty, then a ( ,b ), b ( a, ), and ( ,b ) ( a, ) = , and so X is Hausdorff. If A is nonempty, then a ( ,x ), b ( x, ), and ( ,x ) ( x, ) = for any x A , and therefore, X is Hausdorff. So in particular, single points in X are closed. Suppose now that x X , and A is a closed set, disjoint from x . Then, there exists a basis element ( a,b ) containing x which is disjoint from A . So pick any a ( a,x ), and let U 1 = ( ,a ), V 1 = ( a , ). If no such a exists, then let U 1 = ( ,x ), V 1 = ( a, ). Exactly as before, in both cases, the pair of sets is disjoint....
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This note was uploaded on 05/07/2010 for the course MAT 325 taught by Professor Zoltánszabó during the Fall '09 term at Princeton.
 Fall '09
 ZoltánSzabó
 Topology

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