# Prove that $\sup(S-T)=\sup S-\inf T$

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 $$(1)\,\alpha-\epsilon And by adding $$(1)$$ and $$(2)$$, we obtain $$\alpha-\beta-2\epsilon\leq\alpha-\beta Given that $$x-y\in S-T$$ then $$\alpha-\beta Since $$\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?

• Could you explain how you got $$\alpha-\epsilon<\alpha+y-\beta<x?$$ Oct 7, 2018 at 23:45
• @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? Oct 7, 2018 at 23:56
• Surely $-\epsilon < \beta - y$, so $\alpha - \epsilon < \alpha + \beta - y$? Oct 7, 2018 at 23:57
• It seems a correct implication to me... Oct 7, 2018 at 23:59
• But I got something different from you. I don't think yours is correct. Oct 8, 2018 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 $$\alpha-\epsilon and $$\alpha-\epsilon<\alpha+\beta-y$$. This doesn't tell us $$\alpha+\beta-y, but we don't need that anyways.

• $\alpha+\beta-y<x$ doesn't help me at all because my goal is to show that $\gamma\leq\alpha-\beta$ and $\alpha-\beta\leq\gamma$ implies thaat $\gamma=\alpha-\beta$ Oct 8, 2018 at 0:06
• You used it in your derivation of $\alpha-\beta<x-y$. Oct 8, 2018 at 0:09
• Is my correction valid now? Oct 8, 2018 at 0:14
• 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. Oct 8, 2018 at 0:19
• Consider $\gamma<\gamma+2\epsilon$ for all $\epsilon>0$, but $\gamma\not<\gamma$. The inequality is not strict. This implies $\alpha-\beta\leq\gamma$. Oct 8, 2018 at 0:24