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Material Type: Assignment; Class: Real Variables I; Subject: MATH Mathematics; University: University of Memphis; Term: Fall 2004;
Typology: Assignments
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Due September 16.
i=
j=
Ai,j =
(ai)
i=
Ai,ai
where the second union is over all sequences (ai)โ i=0 of natural numbers.
i=
j=
Ai,j =
(ai)
i=
Ai,ai
where the second union is over all sequences (ai)โ i=0 of natural numbers.
Assume x โ
i=
j=0 Ai,j^. Then, for all^ i,^ x^ โ^
j=0 Ai,j^. Hence, if we fix^ i, there is a j such that x โ Ai,j. Set ai to be the smallest such j. Now x โ Ai,ai for all i, so x โ
i=0 Ai,ai^. In particular,^ x^ โ^
(ai)
i=0 Ai,ai^.
Now assume x โ
(ai)
i=0 Ai,ai.^ Then there exists a sequence (ai)
โ i=0 such that x โ
i=0 Ai,ai , so^ x^ โ^ Ai,ai for all^ i.^ But then, for all^ i,^ x^ โ^
j Ai,j^.^ Hence x โ
i=
j=0 Ai,j^.
By Extensionality, the two sets are equal.
Let ฯ(C) be the ฯ-algebra generated by C and let ฯ(A) be the ฯ-algebra generated by A. Since any ฯ-algebra is also an algebra, ฯ(C) is an algebra containing C, and hence contains A, the smallest algebra containing C. Now ฯ(C) is a ฯ-algebra containing A, so contains ฯ(A), the smallest such ฯ-algebra. Similarly ฯ(A) is a ฯ-algebra that contains A, and hence C, and so contains ฯ(C). Thus ฯ(C) = ฯ(A).
Let X = Z โช {?} with the ordering given by the usual ordering on Z,? โค ?, but? unrelated to any element of Z. Since the usual ordering on Z is a partial ordering, we only need to check the axioms involving ?, which are trivial. Now? is clearly minimal since x โค? implies x = ?. On the other hand, no x โ Z is minimal since x โ 1 < x. Finally, X has no smallest element, since the minimal element? is not โค every other element (or indeed any other element).