Given set $A=\left\{\frac{1}{n}-(-1)^n|n\in\mathbb{N}\right\}$. Calculate $\operatorname{inf}A,\operatorname{sup}A, \operatorname{min}A,\operatorname{max}A$ (if they exist)
attempt:
For all $n\in\mathbb{N}$ $$-1<0-1<\frac{1}{n}-1\leq\frac{1}{n}-(-1)^n\leq\frac{1}{n}+1\leq 1+1=2$$
Showing $\operatorname{inf}A=-1$ which means $\forall \epsilon>0 \:\exists a\in A \,(\operatorname{inf}A+\epsilon >a)$
($n$ is even) There exists an even $n_0>\frac{1}{\epsilon} \iff ... \iff -1+\epsilon>\frac{1}{n_0}-1\in A$
(*)($n$ is odd) that's where I struggle finding some $r$ s.t. $-1+\epsilon>r>\frac{1}{n_0}+1$
Showing $\operatorname{sup}A=2$ which means $\forall \epsilon>0\:\exists a\in A\,(\operatorname{sup}A-\epsilon<a)$
exactly the same struggle as (*)
I am going to use the result that $$\sup (X \cup Y)= \max (\sup (X), \sup(Y))$$ $$\inf (X \cup Y)= \min (\inf(X), \inf(Y))$$
Your given set $A$ can be written as,
$$A=\left\{\frac{1}{n}+1: for \ odd \ n\right\}\bigcup \left\{\frac{1}{n}-1: for \ even \ n\right\}$$
Let $B= \left\{\frac{1}{n}+1: for \ odd \ n\right\}$ and $C= \left\{\frac{1}{n}-1: for \ even \ n\right\} $ which we can also write as
$B= \left\{\frac{1}{2k-1}+1: for \ k\in \mathbb N\right\}$ and $C= \left\{\frac{1}{2k}-1: for \ k \in \mathbb N\right\} $
Finding supremum and infimum of $B$ and $C$ is easy task.