I need to find the supremum and infimum of the following set: $$\Big\lbrace\frac{k-n}{kn}\text{, where k, n }\epsilon\Bbb{N}\Big\rbrace$$ I'm guessing that the supremum is $1$ and infimum is $-1$, but I neither am sure nor know how to prove it.
Supremum and infimum of the following set.
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analysis
supremum-and-infimum
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0Hint: get lim for fixed $k$ , then get lim for fixed $n$ – 2017-01-28
1 Answers
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Since $\color{blue}{\sup\{x+y:x\in A, y\in B\}=\sup\{x:x\in A\}+\sup\{y:y\in B\}}$ $$\sup\Big\lbrace\frac{k-n}{kn}\text{, where k, n }\epsilon\Bbb{N}\Big\rbrace= \sup\Big\lbrace\frac{1}{n}-\frac{1}{k}\text{, where k, n }\epsilon\Bbb{N}\Big\rbrace= \sup\Big\lbrace\frac{1}{n}\text{, where k, n }\epsilon\Bbb{N}\Big\rbrace-\inf\Big\lbrace\frac{1}{k}\text{, where k, n }\epsilon\Bbb{N}\Big\rbrace=1-0=1$$ Similar way shows $\inf=-1$.
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0okay, but why? Where does that come from? @MyGlasses – 2017-01-28