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Boolean subalgebra

Error message

  • Notice: Trying to get property of non-object in question_unanswered_block() (line 404 of /home/jcorneli/beta/sites/all/modules/question/question.module).
  • Notice: Trying to get property of non-object in question_unanswered_block() (line 404 of /home/jcorneli/beta/sites/all/modules/question/question.module).
  • Notice: Trying to get property of non-object in question_unanswered_block() (line 404 of /home/jcorneli/beta/sites/all/modules/question/question.module).
Defines: 
dense Boolean subalgebra
Synonym: 
dense subalgebra
Type of Math Object: 
Definition
Major Section: 
Reference
Groups audience: 

Mathematics Subject Classification

06E05 no label found03G05 no label found06B20 no label found03G10 no label found

Comments

Boolean subalgebra

Let A=P(N)APNA=P(N) be the algebra of subsets of the naturals. Let BBB be the subalgebra of finite and cofinite sets, that is, xAxAx\in A iff |x|<OR|x|<formulae-sequencexORsuperscriptxnormal-′|x|<\infty\ OR\ |x^{{\prime}}|<\infty. Then BBB is dense since every subset contains a finite subset. But it is not true that (xy)(zB):xzyfragmentsfragmentsnormal-(for-allxynormal-)fragmentsnormal-(zBnormal-)normal-:xzy(\forall x<=y)(\exists z\in B):\ x<=z<=y. For example, take y=2N,x=6Nformulae-sequencey2Nx6Ny=2N,\ x=6N (i.e the set of naturals divisible by 2 and the set of naturals divisible by 6). Then xyxyx<=y but no finite or cofinite subset can be between them because only infinite subsets with infinite compliment can be between these two sets.

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