# If $A$ is infinite and bounded, the infimum of the difference set of $A$ is zero.

Let $A$ be a non-empty subset of $\mathbb{R}$. Define the difference set to be

$A_d := \{b-a\;|\;a,b \in A \text{ and } a < b \}$

If $A$ is infinite and bounded then $\inf{A_d} = 0$.

Since $a < b$ we have $b - a > 0$. Thus zero is a lower bound for $A_d$ and $\inf(A_d) \geq 0$. I then want to show that if $\inf(A_d) = \epsilon > 0$ and $A$ is bounded, then $A$ is finite.

Let $\inf(A) = \beta$ and $\sup(A) = \alpha$. Then there can be at most $\lfloor \frac{\alpha - \beta}{\epsilon} \rfloor$ real numbers in $A$. Suppose that there are greater than $\lfloor \frac{\alpha - \beta}{\epsilon} \rfloor + 1$ numbers in $A$. Since $b - a > \epsilon$ for each $a , b \in A$, We have $\alpha > (\lfloor \frac{\alpha - \beta}{\epsilon} \rfloor + 1)(\epsilon) + \beta$. However this is a contradiction, since $(\lfloor \frac{\alpha - \beta}{\epsilon} \rfloor + 1)(\epsilon) + \beta > (\lfloor \frac{\alpha - \beta}{\epsilon} \rfloor)(\epsilon) + \beta\geq ( \frac{\alpha - \beta}{\epsilon} )(\epsilon) + \beta = \alpha$. Thus the cardinality of $A$ must be less than or equal to $\lfloor \frac{\alpha - \beta}{\epsilon} \rfloor + 1$ and thus finite. We have show that if $\inf($A_d$) > 0$ and $A$ is bounded then, $A$ cannot be infinite.

One question I have is whether this would be enough to prove the theorem. I'm sure that there are more effecient ways to formulate the above argument. I feel like this is a good opportunity for the pigeon hole principle but I don't really know how to "invoke" it. Critique is welcomed and appreciated.

• In your definition for $A_d$, are you sure you want $a<b$? For then $a-b<0$ and the infimum wouldn't be $0$. – Clayton Jul 23 '16 at 6:48
• ($b-a$ in the second line for the definition of $A_d$). I think that your proof is correct. For the pigeon hole principe, you can do as follows: let $n$ be a large integer. We have with $x_k=\beta+\frac{k}{n}(\alpha-\beta)$: $$A\subset [\beta, \alpha]=(\cup_{k=0}^{n-1}[x_k, x_{k+1}[)\cup \{\alpha\}$$ As these subsets are disjoint, there must exists two distincts elements of $A$ in the same interval. It is easy to finish. (but this is essentially your proof) – Kelenner Jul 23 '16 at 6:57

Your argument is ok for me. If you want to apply the Pigeon-Hole Principle: We have $A\subset [\inf A, \sup A]=[x,y]$ with $x<y$. For any $r>0$ take $n\in N$ such that $(y-x)/n<r.$ The set of $n$ intervals $S= \{[x+j(y-x)/n,x+(j+1)(y-x)/n] : 0\leq j<n\}$ covers $[x,y].$ Take any set $B$ of $n+1$ members of $A.$ At least two distinct $c,d\in B$ belong to the same member of $S.$ So $\exists c,d\in A\;(0<|c-d|\leq (y-x)/n<r).$

• This is the same as my "simpler proof" -- just using $n$ instead of $N$ and $c, d$ instead of $a, b$ ... – Benjamin Dickman Jul 23 '16 at 16:48
• @Benjamin Dickman And this is also the same as my comment..... – Kelenner Jul 25 '16 at 6:19
• @Kelenner Ha! Good point; okay, up votes all around – Benjamin Dickman Jul 25 '16 at 14:52

Fact: If $A \subset \mathbb{R}$ is infinite and bounded, then it has a limit point.

Accepting this fact, call the limit point $L$ and let $\varepsilon > 0$ be arbitrary.

Since $L$ is a limit point of $A$, the set must contain points arbitrarily close to $L$; in particular, we can find two points $b < a$ in $A$ each within $\varepsilon/2$ of $L$, so that $a - b < \varepsilon$.

Since this can be satisfied for any $\varepsilon > 0$, the infimum of the set $A_d$ is $0$ as desired.

The name of this fact is the Bolzano-Weierstrass Theorem.

A simpler proof: Let $\varepsilon > 0$ be arbitrary. By the Archimedean property of the real numbers, there is some natural number $N$ for which $1/N < \varepsilon$. Divide your bounded set $A$ into $N$ intervals of the same length; since there are infinitely many points, consider any $N+1$ of them: By the pigeonhole principle, some interval must contain two points $b < a$ each in $A$, so that $a - b \leq 1/N < \varepsilon$.

Since $\varepsilon > 0$ was arbitrary, the infimum of the set $A_d$ is $0$ as desired.