Answer to Question #177376 in Discrete Mathematics for Zain Ul Abideen Khan

Question #177376

Find  and  if for every positive integer ,

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


1
Expert's answer
2021-04-15T07:21:17-0400

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 xRx\in R there exists iRi\in R such that xi|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+AiA1\bigcap\limits_{i=1}^{+\infty}A_i\subset A_1 and A1AiA_1\subset A_i for all i, which implies that A1i=1+AiA_1\subset\bigcap\limits_{i=1}^{+\infty}A_i


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