Answer to Question #124174 in Real Analysis for ibrahim sabah

Question

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

Answer 1

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.