Given that
A:={n(−1)n:n∈N} i). Minima and maxima does not exist,suppose on the contrary minima exist when n=2m+1 for some m∈N and it is 2m+11 but 2m+31<2m+11 ,hence contradiction.
Similarly,suppose maxima exist at n=2M and it is 2M but 2M<2M+2 ,hence contradiction.
ii).suppose on the contradiction that A is bounded above,thus by completeness axiom's there exist a least upper bound , say m∈R thus, x≤m,∀x∈A
One can always find ϵ>0 such ϵ=(m−x)/2 such that m−ϵ is not upper bound, thus m−ϵ∈A⟹m−ϵ=p for some p∈A
but
Case I: if n=p is odd then it is trivially not true.
Case II: if n=p ,and n is even,then p+2∈A
but,p+2=2+p/2+ϵ∈/A ,hence contradiction.
iii).Now, observe that
infA=inf{n1:nis odd natural number} And, we already know that inf{1/n:nis odd number}=0
Hence, we are done.
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