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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 $A$

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  • $\begingroup$ The correct definition of $B$ is most probably $B:=\left\{ \frac{n}{n+1}-a\mid a\in A,n\in\mathbb{N}\right\} $. $\endgroup$
    – drhab
    Jan 28, 2019 at 10:17

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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 $B$?

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  • $\begingroup$ Great answer connecting the general and the specific and showing me various ways to do something :) $\endgroup$
    – user459879
    Jan 28, 2019 at 9:54
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    $\begingroup$ 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. $\endgroup$ Jan 28, 2019 at 10:14
  • $\begingroup$ Thanks for empowering me :) $\endgroup$
    – user459879
    Jan 28, 2019 at 10:27
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    $\begingroup$ Aston.As usual, a pleasure to read. $\endgroup$ Jan 30, 2019 at 8:54
  • $\begingroup$ @PeterSzilas Thank you for the compliment! $\endgroup$ Feb 2, 2019 at 8:27
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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:$$\inf B=1-\sup A$$

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