Prove that $(1/n,2/n)$ is an open cover of $(0,1)$

Question

Let $F$ the collection of the open intervals $(1/n,2/n)$, $n\geq 2$. Show that $F$ is an open cover of $(0,1)$.

I can get $n,m\in\mathbb{R}$ such that if $0<x<1$, then $1/n<x<1/m$ but i am having problem in given the form $(1/n,2/n)$.

Thank you

You want an $n$ with $1/n<x<2/n$, that is $n/2<1/x<n$. Can you do that, noting that $1/x>0$? — Angina Seng, Jul 04 '17 at 06:51

score 2 · Answer 1 · answered Jul 04 '17 at 06:59

As @LordSharktheUnknown pointed out, for an arbitrary $x \in (0,1)$ you need to find $n$ such that $\frac 1 n \lt x \lt \frac 2 n$, equivalent $\frac n 2 \lt \frac 1 x \lt n$. Just take $n= \lfloor \frac {1}{x} \rfloor + 1 $

Prove that $(1/n,2/n)$ is an open cover of $(0,1)$

1 Answers1