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I have to show that:

$$\bigcup_{i=2}^\infty [\frac{1}{n},\frac{n-1}{n}] = (0,1)$$

The first part:

$$\bigcup_{i=2}^\infty [\frac{1}{n},\frac{n-1}{n}] \subset (0,1)$$

is easy to show, but for the second part:

$$(0,1) \subset \bigcup_{i=2}^\infty [\frac{1}{n},\frac{n-1}{n}]$$

I don't have any idea how to prove it. I can't use the Principle of Nested Intervals because it's clear that if $I_{k}=[\frac{1}{k},\frac{k-1}{k}]$ where $k$ is a natural number, $I_{k+1} \not\subset I_{k}$ and I have the union.

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Given $x \in (0,1)$, choose a natural number $n > \text{max}\{2,\frac{1}{x},\frac{1}{1-x}\}$; that number exists by the Archmidean principle.

It follows that $x > \frac{1}{n}$ and that $x < \frac{n-1}{n}$ and so $x \in [\frac{1}{n},\frac{n-1}{n}]$.

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  • $\begingroup$ the $n$ that you choose is for $n \ge 2$ ? or it follows by $n>max{ \{\frac{1}{x},\frac{x}{1-x}}\};$ ? $\endgroup$ – angie duque May 30 at 19:52
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    $\begingroup$ I threw a $2$ into the formula for $n$, to cover that necessity. $\endgroup$ – Lee Mosher May 30 at 20:09
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The key things are that

$$\begin{cases} \lim_{n \to \infty} \frac{1}{n} &=0\\ \lim_{n \to \infty} \frac{n-1}{n} &=1 \end{cases}$$

Therefore any point in $(0,1)$ is included in a closed interval $[1/n, {n-1}/n]$ for $n$ properly chosen.

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  • $\begingroup$ (I inadvertently downvoted your post earlier, despite your good answer. The only way I could undo the downvote was via a minor edit, to restore your upvote count.) $\endgroup$ – amWhy May 31 at 16:44
  • $\begingroup$ Ok thanks for that. Normaly if you click again thé downvote arrow, it cancels thé downvote. $\endgroup$ – mathcounterexamples.net May 31 at 16:52
  • $\begingroup$ Not after a few minutes of the downvote. Try it out somewhere where you may have downvoted (or upvoted) a post, more than a few minutes ago. Typically, after a short span of time, unless there is an edit, one cannot change the vote. Anyway, +1 for the trouble I caused you! $\endgroup$ – amWhy May 31 at 16:55
  • $\begingroup$ Thanks. No worries anyhow! $\endgroup$ – mathcounterexamples.net Jun 1 at 8:27

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