S31 - 1st December 2004 Munkres 31 Ex. 31.1 (Morten...

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1st December 2004 Munkres § 31 Ex. 31.1 (Morten Poulsen). Let a and b be distinct points of X . Note that X is Hausdorff, since X is regular. Thus there exists disjoint open sets A and B such that a A and b B . By lemma 31.1(a) there exists open sets U and V such that a U U A and b V V B. Clearly U V = . Ex. 31.2 (Morten Poulsen). Let A and B be disjoint closed subsets of X . Since X normal there exists disjoint open sets U 0 and U 1 such that A U 0 and B U 1 . By lemma 31.1(b) there exists open sets V 0 and V 1 such that A V 0 V 0 U 0 and B V 1 V 1 U 1 Clearly U V = . Ex. 31.3 (Morten Poulsen). Theorem 1. Every order topology is regular. Proof. Let X be an ordered set. Let x X and let U be a neighborhood of x , may assume U = ( a, b ), -∞ ≤ a < b ≤ ∞ . Set A = ( a, x ) and B = ( x, b ). Using the criterion for regularity in lemma 31.1(b) there are four cases: (1) If u A and v B then x ( u, v ) ( u, v ) [ u, v ] ( a, b ). (2) If A = B = then ( a, b ) = { x } is open and closed, since X Hausdorff, c.f. Ex. 17.10. (3) If A = and v B then x ( a, v ) = [ x, v ) [ x, v ) [ x, v ] ( a, b ). (4) If u A and B = then x ( u, b ) ( u, x ] ( u, x ] [ u, x ] ( a, b ). Thus X is regular. ± Ex. 31.5. The diagonal Δ Y × Y is closed as Y is Hausdorff [Ex 17.13]. The map ( f, g ) : X Y × Y is continuous [Thm 18.4, Thm 19.6] so { x X | f ( x ) = g ( x ) } = ( f, g ) - 1 (Δ) is closed. Ex. 31.6. Let p : X Y be closed continuous surjective map. Then X normal Y normal. For this exercise and the next we shall use the following lemma from [Ex 26.12].
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This note was uploaded on 01/12/2011 for the course MATH 110 taught by Professor Brown during the Fall '08 term at Arizona Western College.

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S31 - 1st December 2004 Munkres 31 Ex. 31.1 (Morten...

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