Let $x \in (0,1).$ How to show that $x \in \bigcup \left(0, \frac{n}{n+1}\right)$? Let $x \in (0,1).$ How to show that $x \in \bigcup \left(0, \frac{n}{n+1}\right)$?
My attempt: It seems to me that $(0,1) \subseteq \left(0, \frac{n}{n+1}\right)$ is a valid argument for some $n \in \mathbb{N}$. And if that is the case, it will be easy to show that $x \in \left(0, \frac{n}{n+1}\right)$. But is that really valid or is there a more convincing logic to use?
 A: In order to show that $x$ is in the union;
$$\displaystyle\bigcup_{n=1}^{\infty}{\left(0,\frac{n}{n+1}\right)},$$
you'll have to show that $x$ is in one of the intervals of the form;
$$\left(0,\frac{n}{n+1}\right).$$
Then, you can say that $x$ is also in their union. So you simply need to find some $n$ such that;
$$0<x<\frac{n}{n+1}.$$
To find such a value of $n,$ let $n$ be a positive integer greater than $\frac{1}{1-x}.$ Then we'll get;
\begin{align*}
& n>\frac{1}{1-x}\\
\implies & n>\frac{1}{1-x}\\
\implies & n(1-x)>1\\
\implies & 1-x>\frac{1}{n}\\
\implies & 1-\frac{1}{n}>x\\
\implies & 1-\frac{1}{n}>x>0\\
\implies & \frac{n-1}{n}>x>0.\\
\end{align*}
So, we know that $x$ is between $0$ and $\frac{n-1}{n},$ and so;
$$x\in\left(0,\frac{n-1}{n}\right)$$
$$\implies x\in\displaystyle\bigcup_{n=1}^{\infty}{\left(0,\frac{n-1}{n}\right)}$$
This then implies that all numbers in the set $(0,1)$ are in the set;
$$\bigcup_{n=1}^{\infty}{\left(0,\frac{n-1}{n}\right)}.$$
So we know that;
$$\left(0,1\right)\subseteq \bigcup_{n=1}^{\infty}{\left(0,\frac{n-1}{n}\right)}.$$
Hope this answers your question :)
A: Since
$$
\mathop {\sup }\limits_{n \in \mathbb N} \left( {\frac{n}{{n + 1}}} \right) = 1
$$
it follows that for each $x \in (0,1)$, from Archimedean Property, there is $n_1 \in \mathbb N$ so that $
x < \frac{n_1}{{n_1 + 1}}
$ thus
$
x \in \left( {0,\frac{{n_1 }}{{n_1  + 1}}} \right)
$
and therefore
$
x \in \bigcup\limits_{n \in N} {\left( {0,\frac{n}{{n + 1}}} \right)} 
$.
Hence
$$
\left( {0,1} \right) \subseteq \bigcup\limits_{n \in \mathbb N} {\left( {0,\frac{n}{{n + 1}}} \right)} 
$$
On the otherwise, since
$$
\left( {0,\frac{n}{{n + 1}}} \right) \subset \left( {0,1} \right)
$$
it follows that
$$
\bigcup\limits_{n \in \mathbb N} {\left( {0,\frac{n}{{n + 1}}} \right)}  \subseteq \left( {0,1} \right)
$$
and the equality follows.
A: Since $\frac{n}{n+1}$ is an increasing function with argument $n$ on $\mathbb{N}^+$, then we have
\begin{align*}
\bigcup_{n=1}\left(0,\frac{n}{n+1}\right)&=\left(0,\lim_{n\to \infty}\frac{n}{n+1}\right)\\
&=(0,1).
\end{align*}
