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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?

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    $\begingroup$ Fix a $x$ and find a $n$ such that $x \in (0, \frac{n}{n + 1})$. $\endgroup$
    – Gareth Ma
    May 11 at 4:52
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    $\begingroup$ 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. $\endgroup$ May 11 at 4:55
  • $\begingroup$ If you observe that $\dfrac{n}{n+1}=1-\dfrac{1}{n+1}$, finding the right way wil be easier. $\endgroup$ May 11 at 5:00

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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 :)

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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.

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  • $\begingroup$ "it follows that for each $x\in(0,1)$..." probably you should add "from Archimedean Property" just in case $\endgroup$ May 11 at 5:14
  • $\begingroup$ You are right! Thank you $\endgroup$ May 11 at 13:17
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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*}

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