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June 8th, 2009, 08:17 PM  #1 
Newbie Joined: Jun 2009 Posts: 3 Thanks: 0  Help needed
Help needed with this question: Let Omega(Universe) be an infinite set, and A(Algebra) the class of all subsets B of Omega such that either B or its complement is finite: (a) show that A is a field. (b) is A a sigmafield?why? Thanks. 
June 9th, 2009, 02:53 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,661 Thanks: 648  Re: Help needed Quote:
(b) answer is no. Let a1, a2, a3, ... be a countable set of points in Omega. The following are members of A, {a1}, {a3}, {a5}, .... However their (countable) union is not. It is an infinite set {a1,a3,a5,..} and its complement is also infinite since it contains {a2,a4,a6,...}.  
June 10th, 2009, 09:42 PM  #3 
Newbie Joined: Jun 2009 Posts: 3 Thanks: 0  Re: Help needed
Hi mathamn, Thank you for taking out time and looking at this query. I worked on your reply and the model you have used is good. I would just like to put forward some thing for part b: Your answer: answer is no. Let a1, a2, a3, ... be a countable set of points in Omega. The following are members of A, {a1}, {a3}, {a5}, .... However their (countable) union is not. It is an infinite set {a1,a3,a5,..} and its complement is also infinite since it contains {a2,a4,a6,...}. My analysis: answer is yes. Let a1, a2, a3, ... be a countable set of points in Omega. The following are members of A, {a1}, {a3}, {a5}, .... However their (countable) union is not. It is an infinite set {a1,a3,a5,..} and its complement is finite (which is countable intersection = 'empty set' now) hence it is a sigma field. [Union(A)]^c=Intersection(A^c) = empty set 
June 11th, 2009, 02:52 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,661 Thanks: 648  Re: Help needed Quote:
The complement is the countable intersection of the complements (Omega  {ai} for i odd) which as I stated contains all ai for i even. In case you haven't solved part a, show that it is true for union or intersection of two sets and then use mathematical induction.  

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