Answer to Question #124174 in Real Analysis for ibrahim sabah

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}:n\\in \\mathbb{N}\\}"

i). Minima and maxima does not exist,suppose on the contrary minima exist when "n=2m+1" for some "m\\in \\mathbb{N}" and it is "\\frac{1}{2m+1}" but "\\frac{1}{2m+3}<\\frac{1}{2m+1}" ,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\\in \\mathbb{R}" thus, "x\\leq m,\\forall x\\in A"

One can always find "\\epsilon>0" such "\\epsilon=(m-x)\/2" such that "m-\\epsilon" is not upper bound, thus "m-\\epsilon\\in A\\implies m-\\epsilon=p" for some "p\\in 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\\in A"

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

iii).Now, observe that

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

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

Hence, we are done.


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