Proving infimum and supremum using epsilon definition.

Question

I came to the conclusion that sup$(A)= 3$ and inf$(A)=2$, but I am stuck at part $(a)$ showing it bounded above and below. How do I solve the rest of $(a)$ and $(b)$ using the epsilon definition.

Answer 1

$\sup A$ is indeed $3$. $A$ has two parts, the points of the form $3-2/(n+1)$, $n \ge 1$, $n\in\mathbb{M}$, and the open interval $(-1,1)$. We have $3>3-2/(n+1)$ for all $n\in\mathbb{N}$ given that $2/(n+1)$ is positive. The points of the form $3-2/(n+1)$ approach $3$ as closely as we wish from below as $n$ becomes large, which one can formally state by saying that "given $\epsilon>0$ we can find a point $3-2/(n+1)$ for some $n$ such that $3-2/(n+1)>3-\epsilon$". Because $3$ is also greater than all points in the open interval $(-1,1)$ it turns out that $3$ is $\sup A$.

Turning now to the infimum, note that all points of the form $3-2/(n+1)$ for natural numbers $n$ not zero are at least $3-2/(1+1)=2$. Thus, the points in the open interval are all lower, and the infimum of the set $A$ is the infimum of the open interval. The infimum is the lower end of the interval, $-1$, since every point in the interval is greater than $-1$ and yet points in the interval can be found as close as one wishes to $-1$.