Proving a union of sets How to prove:
$\cup_n[\frac{1}{n},1] = (0,1]$, where $n \in \mathbb N $
The only thing I'm aware of is that we have to prove both left to right and right to left as I'm dealing with sets, and couldn't find a starting point.
Can anyone help with this?
 A: Just observe that for any positive real $x$, there exists $n\in\Bbb N$ such that $${1\over n}<x$$and that $$0<{1\over n}\quad,\quad \forall n\in\Bbb N$$
A: If $x\in\cup_n[\frac{1}{n},1],$ then $x\in[\frac1n,1]$ for some $n$, so, since $\frac1n>0$, it follows that $x\in(0,1]$.
On the other hand, if $x\in(0,1]$, then $x\in[\frac1n,1]$ for all $n>\lfloor \frac1x\rfloor$, so $x\in \cup_n[\frac{1}{n},1] $.
Here $\lfloor y\rfloor$ is the floor function.
A: Well,  if $x \le 0$ then for any $n>0$ then $\frac 1n > 0$ so $x \le 0 < \frac 1n$ and $x \not\in [\frac 1n, 1]$ for any $[\frac 1n, 1]$ and so $x \not \in \cup_n [\frac 1n, 1]$.
And if $x > 1$ then for any $n>0$ then $\frac 1n \le 1 < x$ so $x\not \in [\frac 1n, 1]$ for any $[\frac 1n, 1]$ and so $x \not \in \cup_n [\frac 1n, 1]$.
ANd if $0 < x \le 1$ then.... here's the heart..... $\frac 1x \ge 1$ and there is and $m\in \mathbb N$ so that $\frac 1x < m$ (the archimedean principal).    So $0< \frac 1m < x \le 1$ so $x \in [\frac 1m, 1]$.  So $x \in \cup_n [\frac 1n, 1]$.
So....... $x \in \cup_n [\frac 1n, 1] \iff 0 < x \le 1 \iff x\in (0,1]$.
So  $\cup_n [\frac 1n, 1] =(0,1]$
