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.
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1What are $A$ and $B$? – gt6989b Dec 25 '18 at 15:18
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2If they are sets, which I assume they are, how do you even calculate $\left| A - B \right|$? – Aniruddha Deshmukh Dec 25 '18 at 15:19
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sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh – Daniel Dec 25 '18 at 17:50
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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.
mechanodroid
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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 $|\sup A-\sup B|=1$.
Ben W
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