Question #270634

⋃ 𝐴𝑖



1 and ⋂ 𝐴𝑖



1


a) 𝐴𝑖 = {𝑖, 𝑖 + 1,𝑖 + 2, … }


b) 𝐴𝑖 = {0,𝑖}


c) 𝐴𝑖 = {−𝑖, − 𝑖 + 1, … , −1, 0, 1, … , 𝑖 − 1, 𝑖}


d) 𝐴𝑖 = {−𝑖, 𝑖}

1
Expert's answer
2021-11-24T10:36:38-0500

Let us find i=1Ai\bigcup\limits_{i=1}^{\infty} A_i and i=1Ai\bigcap\limits_{i=1}^{\infty} A_i for the following sets Ai.A_i.


a) If 𝐴𝑖={𝑖,𝑖+1,𝑖+2,}𝐴_𝑖 = \{𝑖, 𝑖 + 1,𝑖 + 2, … \} then i=1Ai=A1=N\bigcup\limits_{i=1}^{\infty} A_i=A_1=\mathbb N and i=1Ai=.\bigcap\limits_{i=1}^{\infty} A_i=\emptyset.


b) For 𝐴𝑖={0,𝑖}𝐴_𝑖 = \{0,𝑖\} we get that i=1Ai={0,1,2,...}=N0\bigcup\limits_{i=1}^{\infty} A_i=\{0,1,2,...\}=\mathbb N_0 and i=1Ai={0}.\bigcap\limits_{i=1}^{\infty} A_i=\{0\}.


c) If 𝐴𝑖={𝑖,𝑖+1,,1,0,1,,𝑖1,𝑖}𝐴_𝑖 = \{−𝑖, − 𝑖 + 1, … , −1, 0, 1, … , 𝑖 − 1, 𝑖\} then i=1Ai=Z\bigcup\limits_{i=1}^{\infty} A_i=\mathbb Z and i=1Ai=A1={1,0,1}.\bigcap\limits_{i=1}^{\infty} A_i=A_1=\{-1,0,1\}.


d) For 𝐴𝑖={𝑖,𝑖}𝐴_𝑖 = \{−𝑖, 𝑖\} we get that i=1Ai=Z{0}\bigcup\limits_{i=1}^{\infty} A_i=\mathbb Z\setminus\{0\} and i=1Ai=.\bigcap\limits_{i=1}^{\infty} A_i=\emptyset.

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