Does $\sup|A-B| \ge |\sup A -\sup B|$?

Question

I have strong intuition that "yes" is the answer, because the difference between any two points between $A$ and $B$ can be only bigger (or equal) to the difference between two specific points. But I couldn't prove it. Would like to see a prove, or a counter example of course.

Answer 1

The answer is yes, and your intuition is correct.

Let $\varepsilon > 0$ and find $a \in A$ such that $\sup A - a < \varepsilon$.

For any $b \in B$ we have $$\sup|A-B| \ge |a-b| \ge a-b > (\sup A - \varepsilon) - \sup B \ge (\sup A - \sup B) - \varepsilon$$ since $b \le \sup B$. Now $\varepsilon$ was arbitrary so it follows $\sup|A-B| \ge \sup A - \sup B$.

The inequality $\sup|A-B| \ge \sup B - \sup A$ follows by symmetry.

Answer 2

Yes. Find $a\in A$ and $b\in B$ to $\epsilon$-approximate $\sup A$ and $\sup B$. Then $|a-b|$ will approximate $|\sup A-\sup B|$, where $\sup|A-B|\geq|a-b|$. Now let $\epsilon\to 0$.

The reverse inequality does not hold like you had originally also wanted. Let $A=\{-1,0\}$ and $B=\{0,1\}$ so that $\sup|A-B|=2$, whereas $\sup A=0$ and $\sup B=1$ so that $|\sup A-\sup B|=1$.