# 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?

• Fix a $x$ and find a $n$ such that $x \in (0, \frac{n}{n + 1})$. May 11 at 4:52
• There is no (positive integer) $n$ such that $(0,1)$ is contained in $(0,n/(n+1))$, so you're already in trouble from your first sentence. May 11 at 4:55
• If you observe that $\dfrac{n}{n+1}=1-\dfrac{1}{n+1}$, finding the right way wil be easier. May 11 at 5:00

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

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; $$\left(0,1\right)\subseteq \bigcup_{n=1}^{\infty}{\left(0,\frac{n-1}{n}\right)}.$$

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.
• "it follows that for each $x\in(0,1)$..." probably you should add "from Archimedean Property" just in case May 11 at 5:14
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*}