Let $A\subseteq R$ be non-empty and bounded . Prove that $\sup A - \inf A =$ $\inf\lbrace b - a : a, b \in R$ and $ a\leq A \leq b \rbrace$
My attempt : Let $A$ be any nonempty subset of $\mathbb{R} $ then suppose inf A = a and sup A = b. It means $\sup A - \inf A = b-a $ and the given
$= \inf \lbrace b-a, b+\epsilon -a, b+\epsilon - a+\epsilon_1 \rbrace$
$= b-a.$
Hence proved.
Is it correct?
I'm not really sure how what you've written is a proof. It seems like you just assumed what $\inf A$ and $\sup A$ were. I think the problem is to show that $\inf A = a$ and $\sup B = b$, then it is clear that $\sup A - \inf A = b-a$. To show that $a$ is the infimum of $A$, just assume that there is a greater lower bound and argue by contradiction. (It is clear that $a$ is a lower bound). Then do the essentially the same thing with $b$ by assuming there is a smaller upper bound and arguing by contradiction.