# Supremum and infimum of $\{ \frac{n}{n+1}-a | a \in A\}$ where A is a bounded set.

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 $$A$$

• The correct definition of $B$ is most probably $B:=\left\{ \frac{n}{n+1}-a\mid a\in A,n\in\mathbb{N}\right\}$. – drhab Jan 28 '19 at 10:17

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.

Result : For any two subsets $$X$$ and $$Y$$ of the real line, define $$X+Y = \{x + y : x \in X, y \in Y\}$$. If $$X,Y$$ are bounded, then so is $$X+Y$$. Furthermore, we also have the following formulas : $$\inf X + \inf Y = \inf(X+Y)$$, and $$\sup(X+Y) = \sup X + \sup Y$$.

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, $$\sup X+Y = \sup X + \sup Y$$ .

Result : If $$A$$ is bounded, then $$-A$$ is bounded, with $$\sup (-A) = - \inf (A)$$ and $$\inf (-A) = -\sup A$$.

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 $$B$$?

• Great answer connecting the general and the specific and showing me various ways to do something :) – Algebra geek Jan 28 '19 at 9:54
• You are welcome! On the three techniques you mentioned at the end of your question, at least in the starting you will see mostly the first being used : that is, proceeding by definition.In the case of taking the supremum of a sequence, the second condition may be used , and the third is essentially a contradiction of the first, seen more occasionally though. – Teresa Lisbon Jan 28 '19 at 10:14
• Thanks for empowering me :) – Algebra geek Jan 28 '19 at 10:27
• Aston.As usual, a pleasure to read. – Peter Szilas Jan 30 '19 at 8:54
• @PeterSzilas Thank you for the compliment! – Teresa Lisbon Feb 2 '19 at 8:27

If $$B$$ is defined as in your question then $$\inf B=\frac{n}{n+1}-\sup A$$ and $$\sup B=\frac{n}{n+1}-\inf A$$.

Most probably this is about a set $$B$$ defined as: $$B:=\left\{ \frac{n}{n+1}-a\mid a\in A,n\in\mathbb{N}\right\}$$

Indeed from $$a\geq\inf A$$ we can conclude that $$1-\inf A$$ is an upper bound of $$B$$ on base of $$\frac{n}{n+1}-a\leq\frac{n}{n+1}-\inf A\leq1-\inf A$$ Now if $$1-\inf A$$ is not the least upper bound of $$B$$ then there must be a smaller upper bound. So if $$c$$ denotes this smaller upper bound of $$B$$ then we must have $$\frac{n}{n+1}-a\leq c<1-\inf A$$for every $$a\in A$$ and every $$n\in\mathbb N$$.

Now be aware that we can take $$n\in\mathbb N$$ such that $$\frac{n}{n+1}$$ is very very close to $$1$$ and that we can take $$a\in A$$ very very close to $$\inf A$$.

In fact: as close as we want.

Let $$\epsilon>0$$ such that $$1-\inf A-\epsilon>c$$

We can take some $$n\in \mathbb N$$ such that $$\frac{n}{n+1}>1-\frac12\epsilon$$ and $$a\in A$$ such that $$a<\inf A+\frac12\epsilon$$ leading to:$$\frac{n}{n+1}-a>1-\inf A-\epsilon>c$$

This however contradicts that $$c$$ is an upper bound of $$B$$ and allows us to conclude that $$1-\inf A$$ is the least upper bound of $$B$$.

In mathematical language:$$\sup B=1-\inf A$$

Similarly it can be proved that:$$\inf B=1-\sup A$$