# How to prove that $\sup(A-B) = \sup(A) - \inf(B)$?

How to prove that $$\sup(A-B) = \sup(A) - \inf(B)$$?

My attempt: Let $$c \in A-B$$ and define $$c= a-b$$, where $$a \in A$$ and $$b \in B$$. Then $$a-b \leq \sup(A) - \inf(B)$$. Hence, $$\sup(A-B) \leq \sup(A) - \inf(B)$$.

Moreover, for any $$\epsilon >0$$, $$\sup(A) \leq a + \epsilon$$ and $$\inf(B) \geq b - \epsilon$$. This implies that $$\sup(A) + b - \epsilon \leq \inf(B) + a + \epsilon$$ $$\sup(A) - \inf(B) \leq a-b + 2\epsilon \leq \sup(A-B)$$

Is this proof correct or logical?

• How are you arriving at $\sup(A)\leq a+\epsilon$ and $\inf(B)\geq b-\epsilon$? This seems dependent on your choice of $a,b,\epsilon$. Apr 30 at 4:07
– Koro
Apr 30 at 6:04

Let $$z \in A-B, z=a-b$$ for $$a \in A, b \in B$$. Since $$\sup (A)$$ is the least upper bound for A, inf(B) is the greatest lower bound for B, we know $$a \leq \sup (\mathrm{A}), \mathrm{b} \geq \inf (\mathrm{B})$$ so $$z=a-b \leq \sup (A)-\inf (B)$$. Therefore $$\operatorname{sub}(A)$$-inf(B) is an upper bound for A-B.

Fix $$\epsilon>0$$, then there exists $$a \in A$$ such that $$\sup (A)-\frac{\epsilon}{2}, and there exists $$b \in B$$ such that $$\inf (\mathrm{B})-\frac{\epsilon}{2}>\mathrm{b}$$. Let $$k$$ be another upper bound for $$\mathrm{A}-\mathrm{B}$$, then $$a-b \leq k$$ for $$a \in A, b \in B$$. Therefore, $$\begin{array}{c} \sup (A)-\inf (B) Therefore, $$\sup (A)-\inf (B)=\sup (A-B)$$.

– Sonu
May 3 at 8:22

$$A-B=\{a-b : a\in A,b\in B\}$$

Then $$\forall a\in A, b\in B$$

$$a\le \sup(A)$$ and $$b\ge \inf(B)$$

Implies $$a-b\le \sup(A) -\inf(B)$$

Hence, $$\sup(A) -\inf(B)$$ is an upper bound of the set $$A-B$$ and $$\sup(A-B)$$ is the least upper bound.

Implies $$\sup(A-B) \le \sup(A) -\inf(B)$$