Prove that $\bigcup_{n \in \mathbb{N}} (1/n,1-1/n) = (0,1)$ For $\bigcup_{n \in \mathbb{N}} (1/n,1-1/n)  = (0,1)$, I know that I have to prove that: $$\bigcup_{n \in \mathbb{N}} (1/n,1-1/n) \subset (0,1) \tag 1 $$
and 
$$ (0,1) \subset \bigcup_{n \in \mathbb{N}} (1/n,1-1/n) \tag 2$$ but I have problems when I try to prove (2)
 A: First suppose that $x\in(0,1)$. Then set $d_1=|x-0|$ and $d_2=|x-1|$. If we set $D=\min\{d_1,d_2\}$ then there is an integer $n$ such that $1/n<D$ By the Archimedean property. What can you say about $(1/n,1-1/n)$?
A: Suppose $x\in(0,1).$ To show that $x\in\bigcup\limits_{n=1}^\infty \left( \frac 1 n, 1-\frac 1 n \right),$ you need to show that there is some natural number $n$ for which $x\in\left( \frac 1 n, 1 - \frac 1 n \right),$ or equivalently, $x$ and $1-x$ are both less than $1/n$.
Suppose there is no such natural number $n.$ Then for every natural number $n,$ we have $x>1/n,$ so $\dfrac 1 x>n,$ and similarly $\dfrac 1 {1-x} > n.$ That means the set of all natural numbers has an upper bound, and hence a smallest upper bound. Let $c$ be that smallest upper bound. Then $c-1,$ being smaller than $c,$ is not an upper bound of $\mathbb N.$ So for some $n\in\mathbb N,$ $c-1\ge n.$ Hence $c\ge n+1,$ so $c$ is not an upper bound of $\mathbb N.$ There your have a contradiction.
A: Here is how I would write down a straightforward direct proof.$%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\openint}[2]{\left(#1,#2\right)}
\newcommand{\max}{\mathbin{\text{max}}}
%$
Letting $\;n\;$ range over the positive integers, we calculate which $\;x\;$ are in the left hand side set:
$$\calc
    x \in \bigcup_n \openint{\tfrac 1 n}{1 - \tfrac 1 n}
\op\equiv\hint{definition of $\;\bigcup\;$}
    \langle \exists n :: x \in \openint{\tfrac 1 n}{1 - \tfrac 1 n} \rangle
\op\equiv\hint{definition of open interval}
    \langle \exists n :: x \in \mathbb R \;\land\; \tfrac 1 n \le x \le 1 - \tfrac 1 n \rangle
\op\equiv\hints{the inequalities imply $\;0<x\;$ and $\;x<1\;$, because $\;n > 0\;$}\hints{-- working towards the goal, and this will help manipulate}\hint{the inequalities below}
    \langle \exists n :: x \in \mathbb R \land 0<x<1 \;\land\; \tfrac 1 n \le x \le 1 - \tfrac 1 n \rangle
\op\equiv\hints{definition of open interval; move out of $\;\exists n\;$}\hint{-- introduce our goal}
    x \in \openint 0 1 \;\land\; \langle \exists n :: \tfrac 1 n \le x \le 1 - \tfrac 1 n \rangle
\op\equiv\hints{isolate $\;n\;$ in left inequality using $\;n>0\;$ and $\;x>0\;$;}\hint{isolate $\;n\;$ in right inequality using $\;n>0\;$ and $\;x<1\;$}
    x \in \openint 0 1 \;\land\; \langle \exists n :: n \ge \tfrac 1 x \;\land\; n \ge \tfrac 1 {1-x} \rangle
\op\equiv\hints{second part is true for, e.g., $\;n := \left\lceil\tfrac 1 x \mathbin{\text{max}} \tfrac 1 {1-x}\right\rceil\;$,}\hint{which is a positive integer}
    x \in \openint 0 1
\endcalc$$
By set extensionality this completes the proof.
A nice aspect of this proof is that your $\Ref{1}$ and $\Ref{2}$ are proven together.  Also, note how this proof starts at the most complex side, and then is driven by expanding definitions and simplifying.
$%
\endgroup
%$
