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 $A$
Best Answer
For example, take the case where we want to show $\sup B= 1 - \inf A$. We have already shown that $1-\inf A$ is an upper bound, so $1 - \inf A \geq \sup B$.
For the other way, pick $\epsilon > 0$. We want to show that there is some $n$ and some $a \in A$ with $\frac{n}{n+1} - a < 1 - \inf A- \epsilon$.
This is simple : note that we can find $N$ so that $1-\frac{N}{N+1} < \frac\epsilon 2$(take any $N > \frac 2 \epsilon$), and then we can find $a \in A$ such that $a - \inf A < \frac{\epsilon}{2}$ (by the definition of infimum).
Add these up and rearrange to get $\frac{N}{N+1}- a> 1 - \inf A - \epsilon$.
In other words, if $A$ AND $n$ are involved, then split the given $\epsilon$ into smaller $\epsilon$-numerator fractions, obtain separate equations for $A$ and $n$ and then combine them.
I leave you to figure out how the second one can be done. Remember, obtain separate equations for $A$ and $n$ and then combine them.
This is the sort of situation where generality helps.
Proof : I will do it for the supremum, you figure out the infimum : it is exactly the same.
For any $z \in X+Y$, we know $z = x+y$ for some $x\in X,y \in Y$. Therefore $z \leq \sup X + \sup Y$. It follows that $\sup X + \sup Y$ is an upper bound for $X+Y$, so $\sup X+Y \leq \sup X + \sup Y$.
For the other way, fix $\epsilon > 0$. Let $ x',y' $ be such that $\sup X - x' < \frac \epsilon 2$ and $\sup Y - y' < \frac \epsilon 2$. Add and conclude that $(\sup X + \sup Y) - (x'+y') < \epsilon$. Therefore, $\sup X+Y = \sup X + \sup Y$ .
Prove this yourself.
Now, just note for your question that $B = S + (-A)$, where $A$ is some bounded set and $S = \{\frac{n}{n+1} : n \in \mathbb N\} \cup \{0\}$. Can you use the general result to find the infimum and supremum of $B$?