how to prove supremum and infimum

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Here the supremum case: For any positive integers $m,n$ with $m<n$, then $m/n<1$, so we get immediately $\sup X\leq 1$. Now we claim that $\sup X=1$. Given $\epsilon>0$, by Archimedean property, we can find some positive integer $N$ such that $1/N<\epsilon$, then $(N-1)/N\in X$ and satisfies $(N-1)/N=1-1/N>1-\epsilon$, so $\sup X>1-\epsilon$. Since this is true for all $\epsilon>0$, then $\sup X\geq 1$, so we get $\sup X=1$.

Similar argument using Archimedian property will show that $\inf X=0$.

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Moran Tailu
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Updated on August 15, 2020

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  • Moran Tailu
    Moran Tailu about 3 years

    I started to learn this subject, and I have understood what It's mean, but I cant find out how to find and to prove it. for example :

    let $X$ be the set of all rational number in form $m/n$ so $ 0<m<n $ prove that $\inf X = 0$ and $\sup X =1$, and prove that $\max X$ and $\min X$ does not exist.

    how can I solve this? can you please explain it step by step so I would understand