Supremum and infimum of the following set.

Question

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.

Answer 1

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 $\inf=-1$.