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# sol1 - Solutions Problem Set 1 Problem 1(a F must contain...

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Solutions - Problem Set 1 Problem 1 (a) F must contain all intersections, unions, and complements the two sets, so { 2 } , { 1 , 3 , 4 } , { 1 , 2 , 4 } , { 3 } are elements. Now, intersections and unions of these taken together with the original two sets, and their complements, yield { 2 , 3 } , { 1 , 4 } , { 1 , 2 , 3 } , { 4 } , { 1 } At this point we have all elements ( { 1 } , { 2 } , { 3 } , { 4 } ), so F will be the power set of Ω. (b) F must contain all intersections, unions, and complements the two sets, so F = {{ 1 , 2 } , { 3 , 4 } , { 1 , 2 , 3 , 4 } , φ } Problem 2 (a) One possible choice is Ω = ZZ + ∪ { 0 } . (b) F can be the power set of Ω. (c) A 1 = { t < 1 } and A 2 = { t > 2 } (d) B 1 = { t < 2 } and B 2 = { t > 1 } Problem 3 (a) E k & since { N 1 > k } ⊃ { N 1 > k 0 } for k 0 > k . (b) None. (c) E k % since { min m 6 k N m = 0 } implies the occurance of { min m 6 k 0 N m = 0 } for k 0 > k and also if { min m 6 k N m = 0 } does not occur, { min m 6 k 0 N m = 0 } can still occur for some k 0 > k . (d) None. Consider the sequence N 1 = 1 , N 2 = 0 , N 3 = 3 for which the sequence of events goes from occuring ( E 1 occurs) to not occuring ( E 2 does not occur) and back to occuring again ( E 3 occurs).

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sol1 - Solutions Problem Set 1 Problem 1(a F must contain...

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