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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $$\sup (A - B) = \alpha - \beta.$$

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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.

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  • $\begingroup$ $\alpha+\beta-y<x$ 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
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    $\begingroup$ 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

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