text: # 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.
explanation: The provided text is a detailed mathematical discussion that includes a theorem statement, a proof, and a critique of the proof. The main theorem is about the infimum of the difference set of an infinite and bounded subset of the real numbers. The text also includes a question from the author asking for feedback on the proof and whether it is sufficient to prove the theorem. Additionally, there are comments from other users suggesting alternative proofs and pointing out a potential issue in the definition of the difference set.
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distilabel_metadata.role: system
distilabel_metadata.content: # 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed mathematical discussion that includes a theorem statement, a proof, and a critique of the proof. The main theorem is about the infimum of the difference set of an infinite and bounded subset of the real numbers. The text also includes a question from the author asking for feedback on the proof and whether it is sufficient to prove the theorem. Additionally, there are comments from other users suggesting alternative proofs and pointing out a potential issue in the definition of the difference set.
  
  Label: question-and-answer