Let $S$ and $T$ be nonempty sets of real numbers and define $$S-T=\{s-t|s\in S,t\in T\}$$ Show that if S and T are bounded then $$\sup(S-T)=\sup S-\inf T\\ \inf(S-T)=\inf S-\sup T.$$

My proof:

Since $S,T\subset\mathbb{R}$ are nonempty and bounded, then, by the Completeness Axiom, we have $\alpha=\sup S,\beta=\inf T$. Let $m\in S-T$ so that $m=s-t\leq\alpha-\beta\,$ which implies that $S-T$ is bounded above so a supremum exists: denote $\gamma=\sup(S-T)\leq\alpha-\beta$. By a theorem stated in my book, pick any $\epsilon>0$, then $\exists x\in S, \exists y\in T$ such that $$(1)\,\alpha-\epsilon<x,\quad\quad(2)\,y<\beta+\epsilon\equiv-\beta-\epsilon<-y.$$ And by adding $(1)$ and $(2)$, we obtain $$\alpha-\beta-2\epsilon\leq\alpha-\beta<x-y$$ Given that $x-y\in S-T$ then $$\alpha-\beta<x-y\leq\gamma.$$ Since $\gamma\leq\alpha-\beta$ and $\alpha-\beta\leq \gamma$ we have that $\gamma=\alpha-\beta$. $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square$$

Would this work?

  • $\begingroup$ Could you explain how you got $$\alpha-\epsilon<\alpha+y-\beta<x?$$ $\endgroup$ – Theo Bendit Oct 7 '18 at 23:45
  • $\begingroup$ @TheoBendit By inequality $(2)$ we have that $\epsilon>y-\beta$ and using $(1)$ we have that $\alpha-\epsilon<\alpha+y-\beta$. Would this implication be correct? $\endgroup$ – Mauritius87 Oct 7 '18 at 23:56
  • 1
    $\begingroup$ Surely $-\epsilon < \beta - y$, so $\alpha - \epsilon < \alpha + \beta - y$? $\endgroup$ – Theo Bendit Oct 7 '18 at 23:57
  • $\begingroup$ It seems a correct implication to me... $\endgroup$ – Mauritius87 Oct 7 '18 at 23:59
  • $\begingroup$ But I got something different from you. I don't think yours is correct. $\endgroup$ – Theo Bendit Oct 8 '18 at 0:00

You fix $\varepsilon > 0$ and choose $x$ and $y$ such that $\alpha - \varepsilon < x \le \alpha$ and $-\beta - \varepsilon < -y \le -\beta$. Choosing such $x$ and $y$ represent a small concession: you know that $\alpha$ and $-\beta$ may not be achievable, but you know that you can get as close as you want to these bounds, and you're happy to concede $\varepsilon$ distance from these ideals. That's why you're not going to be able to cancel these $\varepsilon$s: you can't make two of these concessions, and expect anything except that these concessions will add to each other.

Instead, observe that, $$\alpha - \beta - 2 \varepsilon < x - y \le \alpha - \beta.$$ It follows that $$\alpha - \beta - 2\varepsilon < \sup (A - B) \le \alpha - \beta.$$ But, there is only one number that satisfies this for any $\varepsilon > 0$. We must have $$\sup (A - B) = \alpha - \beta.$$


From (2) we get $-\epsilon<\beta-y$, so $\alpha-\epsilon<\alpha+\beta-y$. The proof needs a little work, but you can salvage it. You don't need to use (1).

Also, we have $\alpha-\epsilon<x$ and $\alpha-\epsilon<\alpha+\beta-y$. This doesn't tell us $\alpha+\beta-y<x$, but we don't need that anyways.

  • $\begingroup$ $\alpha+\beta-y<x$ doesn't help me at all because my goal is to show that $\gamma\leq\alpha-\beta$ and $\alpha-\beta\leq\gamma$ implies thaat $\gamma=\alpha-\beta$ $\endgroup$ – Mauritius87 Oct 8 '18 at 0:06
  • $\begingroup$ You used it in your derivation of $\alpha-\beta<x-y$. $\endgroup$ – Melody Oct 8 '18 at 0:09
  • $\begingroup$ Is my correction valid now? $\endgroup$ – Mauritius87 Oct 8 '18 at 0:14
  • $\begingroup$ When you write $\alpha-\beta-2\epsilon\leq\alpha-\beta<x-y$, I don't think that follows. This would lead to a contradiction, for $\gamma\leq\alpha-\beta$ and $\gamma>\alpha-\beta$. By trichotomy this cannot happen. You can just use that $\alpha-\beta-2\epsilon<x-y$, therefore $\alpha-\beta\leq x-y$. This leads to your desired result. $\endgroup$ – Melody Oct 8 '18 at 0:19
  • 1
    $\begingroup$ Consider $\gamma<\gamma+2\epsilon$ for all $\epsilon>0$, but $\gamma\not<\gamma$. The inequality is not strict. This implies $\alpha-\beta\leq\gamma$. $\endgroup$ – Melody Oct 8 '18 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.