I tried to prove: $\inf(A-B) = \inf(A) – \sup(B)$. I was hoping someone could check my proof and perhaps give me some tips, I didn't quite get the ending right. Thanks in advance!
First, I show that $\inf(A-B) = \inf(A) + \inf(-B)$:
Suppose $x \in A$ and $y \in B$. Then $-y \in -B$.
$x \geq \inf(A)$ and $-y \geq \inf(-B)$. This gives us $x-y \geq \inf(A) + \inf(-B)$. So $\inf(A) + \inf(-B)$ is a lower-bound and $\inf(A-B) \geq \inf(A) + \inf(-B)$
Now, suppose $\epsilon>0$. Then $\exists_{x\in A, -y\in -B}$ such that $x < \inf(A) – \frac{\epsilon}{2}$ and $-y < \inf(B) – \frac{\epsilon}{2}$
So $x-y < \inf(A) + \inf(-B) – \epsilon$. This indicates that $\inf(A) + \inf(-B) \geq \inf(A-B)$.
So $\inf(A) + \inf(-B) = \inf(A-B)$.
So now I have to prove that $\inf(-B) = -\sup(B)$. My attempt:
Suppose $y\in B$. Then $y \leq \sup(B)$ this gives $-y \geq -\sup(B)$.
Also $-y \geq \inf(-B)$ by definition. So $\inf(-B) = -\sup(B)$?
Best Answer
If I'm not mistaken, the first portion is correct. Concerning the inf(−B) = −sup(B) portion, consider -sup(B). If you can show that -sup(B) is a lower bound of -B, perhaps by first considering sup(B), the solution should follow by combining a similar result on the -inf(-B) being an upper bound of B, i.e., eventually showing -sup(B) $\leqslant$ inf(-B) and -sup(B) $\geqslant$ inf(-B). Note my work assumes B is a nonempty bounded subset of R.