This is a pretty standard question, but I am having trouble to complete the final step.
I am asked to find the supremum and infimum of $B=\{ \frac{n}{n+1}-a | a \in A, n \in \mathbb N\}$ where A is a bounded set, and $\mathbb N$ includes $0$.
A first guess would be to find a good upper bound. Notice that: $a \geq \inf(A) \implies -a \leq \inf(A) $ and that $ 0\leq\frac{n}{n+1}< 1$
so we get that:
$$ \frac{n}{n+1}-a \leq 1-\inf(A).$$ and hence we have found an upper bound for $B$, which could be a candidate for the supremum. Similarly we can derive that: $$ \frac{n}{n+1}-a \geq 0-\sup(A).$$
This is a candidate for the infimum of $B$.
Now how would I prove that: $$ \sup(B)=1-\inf(A) $$ $$ \inf(B)=-\sup(A) $$
Standard approaches:
1)$ \forall \epsilon>0$ there must exist an element $x \in X$ such that $x>\sup(X) - \epsilon$
2) Bounded and increasing, therefore the sequence converges to supremum. (Monotone convergence theorem)
3) There is no smaller upper bound, we force a contradiction
I do not see how this would work as we have an $n$ AND a set $A$