Question #177376

Find  and  if for every positive integer ,

a)   Ai={0,i}           b)  Ai=[-i,i].


Expert's answer

a)  if for every positive integer i, Ai={0,i}A_i=\{0,i\} , then

⋃i=1+∞Ai=⋃i=1+∞{0,i}={0}∪⋃i=1+∞{i}={0}∪N\bigcup\limits_{i=1}^{+\infty}A_i=\bigcup\limits_{i=1}^{+\infty}\{0,i\}=\{0\}\cup \bigcup\limits_{i=1}^{+\infty}\{i\}=\{0\}\cup N


⋂i=1+∞Ai=⋂i=1+∞{0,i}={0}\bigcap\limits_{i=1}^{+\infty}A_i=\bigcap\limits_{i=1}^{+\infty}\{0,i\}=\{0\}

 b) if for every positive integer i, Ai=[−i,i]A_i=[-i,i] , then

⋃i=1+∞Ai=⋃i=1+∞[−i,i]=(−∞,+∞)\bigcup\limits_{i=1}^{+\infty}A_i=\bigcup\limits_{i=1}^{+\infty}[-i,i]=(-\infty,+\infty)

since for every x∈Rx\in R there exists i∈Ri\in R such that ∣x∣≤i|x|\leq i, i.g. x∈[−i,i]x\in [-i,i]


⋂i=1+∞Ai=⋂i=1+∞[−i,i]=A1=[−1,1]\bigcap\limits_{i=1}^{+\infty}A_i=\bigcap\limits_{i=1}^{+\infty}[-i,i]=A_1=[-1,1]

since ⋂i=1+∞Ai⊂A1\bigcap\limits_{i=1}^{+\infty}A_i\subset A_1 and A1⊂AiA_1\subset A_i for all i, which implies that A1⊂⋂i=1+∞AiA_1\subset\bigcap\limits_{i=1}^{+\infty}A_i


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