I'm trying to show that $\bigcup_{n\in \mathbb N^*}[0,1-\dfrac{1}{n}]=[0,1)$ and I'm stuck at the following step:
If $x\in [0,1)$ how to justify that there exists $n\in \mathbb N$ such that $x\leq 1-\dfrac{1}{n}$ ?
It seems like an Archimedean property but I can't see how to use it: $\forall x>0$ and $\forall y>0$ there exists a positive integer $n $ such that $nx>y$
"∀x>0 and ∀y>0 there exists a positive integer n such that nx>y"
Right. So let $x' = 1-x > 0$ and $y = 1$. So there is an $n$ so that $nx' > y$
... or in other words.... $n(1-x) > 1$
... or in other words $(1-x) > \frac 1n$
... or in other words $1-\frac 1n > x$.
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Perhaps more direct. For any $\epsilon > 0$ then there is a $\frac 1n < \epsilon$. So just let $\epsilon = 1- x$. and you get $0 < \frac 1n < 1-x$.
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How do we justify that for any $\epsilon > 0$ there is a $\frac 1n < \epsilon$? Or for that matter how do we justify the Archmedian principal that there is an $n$ so that $nx > y$?
Well.... if $1 > \epsilon = 1 - x > 0$ then $\frac 1{\epsilon} > 1$ and we can find an $n$ so that $n > \frac 1{\epsilon}$ which means $\frac 1n < \epsilon$.
Likewise if $n > \frac yx$ then ... well, ....
We have that
$$x\leq 1-\dfrac{1}{n}\iff \frac1n\le 1-x\iff n\ge \frac{1}{1-x}.$$
So, it is enough to take
$$n=\left[\frac{1}{1-x}\right]+1$$ where $[]$ denotes the integer part.
We have $x \lt 1$, so $1-x\gt0$, so $1/(1-x)\gt0$, so by Archimedian property there is a natural number $n$ such that $1/(1-x)\leq n$, or $1/n\leq 1-x$, or $x\leq 1-1/n$.