TITLE: Basic inequality question
QUESTION [1 upvotes]: I'm trying to find a value of $n$ inside of this intersection of intervals:$\bigcap\limits_{n \in \mathbb{N}}{[\frac{-1}{n},1)}$ such that $y \notin [0,1) \implies y \notin \bigcap\limits_{n \in \mathbb{N}}{[\frac{-1}{n},1)}$. If $y>1$ the answers trivial but if $y < 0$ I am struggling with a basic question.Is this correct way to find a natural number $n$ such that $y<\frac{-1}{n}<0$? 
$y<\frac{-1}{n} \implies \frac{1}{y}>-n \implies n>\frac{-1}{y}$.So choose $n \in \mathbb{N} \ni n>\frac{-1}{y}$ then $y\notin \bigcap\limits_{n \in \mathbb{N}}{[\frac{-1}{n},1)}$.
I know this is a very elementary question but does $y < 0$ mean that when I take the reciprocal of $\frac{1}{y}$ this sign of the inequality does not change?Also is this the correct way to prove the implication above?

REPLY [1 votes]: You are certainly correct that: if for $y<0$ there exists $n\in\mathbb{N}$ such that $y<-\frac{1}{n}<0$, then $y\not\in \bigcap_{n=1}^{\infty}\left[-\frac{1}{n},1\right)$.
For your second question, you can not simply say the inequality does not change when taking the reciprocal since $y<0$. For example, say $y=-\frac{1}{4}$ and $n=5$. Then it is clearly true that
$$-\frac{1}{4}<-\frac{1}{5}<0.$$
However, it is not true that
$$-4<-5.$$
What you might be thinking is that if $y<-\frac{1}{n}$, then $-\frac{1}{y}<n$. From the example above, we would have $4<5$, which is clearly true. Thus, proving the inequality simplifies to showing that if $y<0$, then there exists $n\in\mathbb{N}$ such that $-\frac{1}{y}=\frac{1}{|y|}<n$. However, since $\frac{1}{|y|}$ is a positive real, this is nothing more than the Archimedean Property! Thus, the inequality is true for $n>\frac{1}{|y|}$ and the proposition is proved.
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