Question #124174
Consider the set A={n^(-1)^n:n€N} (i). Find maximum and minimum if there exists. (ii). Show that the set is not bounded above. (iii). Show that Inf A =0
1
Expert's answer
2020-06-29T17:35:15-0400

Given that

A:={n(1)n:nN}A:=\{n^{(-1)^n}:n\in \mathbb{N}\}

i). Minima and maxima does not exist,suppose on the contrary minima exist when n=2m+1n=2m+1 for some mNm\in \mathbb{N} and it is 12m+1\frac{1}{2m+1} but 12m+3<12m+1\frac{1}{2m+3}<\frac{1}{2m+1} ,hence contradiction.

Similarly,suppose maxima exist at n=2Mn=2M and it is 2M2M but 2M<2M+22M<2M+2 ,hence contradiction.


ii).suppose on the contradiction that AA is bounded above,thus by completeness axiom's there exist a least upper bound , say mRm\in \mathbb{R} thus, xm,xAx\leq m,\forall x\in A

One can always find ϵ>0\epsilon>0 such ϵ=(mx)/2\epsilon=(m-x)/2 such that mϵm-\epsilon is not upper bound, thus mϵA    mϵ=pm-\epsilon\in A\implies m-\epsilon=p for some pAp\in A

but

Case I: if n=pn=p is odd then it is trivially not true.


Case II: if n=pn=p ,and nn is even,then p+2Ap+2\in A

but,p+2=2+p/2+ϵAp+2=2+p/2+\epsilon\notin A ,hence contradiction.

iii).Now, observe that

infA=inf{1n:nis odd natural number}\inf A=\inf \{\frac{1}{n}:n \:\text{is odd natural number}\}

And, we already know that inf{1/n:nis odd number}=0\inf \{1/n:n \text{is odd number}\}=0

Hence, we are done.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS