a) if for every positive integer i, Ai={0,i} , then
i=1⋃+∞Ai=i=1⋃+∞{0,i}={0}∪i=1⋃+∞{i}={0}∪N
i=1⋂+∞Ai=i=1⋂+∞{0,i}={0}
b) if for every positive integer i, Ai=[−i,i] , then
i=1⋃+∞Ai=i=1⋃+∞[−i,i]=(−∞,+∞)
since for every x∈R there exists i∈R such that ∣x∣≤i, i.g. x∈[−i,i]
i=1⋂+∞Ai=i=1⋂+∞[−i,i]=A1=[−1,1]
since i=1⋂+∞Ai⊂A1 and A1⊂Ai for all i, which implies that A1⊂i=1⋂+∞Ai
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