# How can I justify that the open interval $(0,1)$ is the infinite union of closed intervals?

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.

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 $$x \in [\frac{1}{n},\frac{n-1}{n}]$$.

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

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.

• (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.) – amWhy May 31 at 16:44
• Ok thanks for that. Normaly if you click again thé downvote arrow, it cancels thé downvote. – mathcounterexamples.net May 31 at 16:52
• 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! – amWhy May 31 at 16:55
• Thanks. No worries anyhow! – mathcounterexamples.net Jun 1 at 8:27