# Identify a countable union of nested intervals using the Archimedean principle [closed]

$\displaystyle\bigcup_{n=2}^\infty \left[\frac1n,3-\frac 2n\right]=(0,3)$. I can't prove using limits. I have to use the Archimedean principle and I don't know how to go about doing that..

## closed as off-topic by Davide Giraudo, anomaly, Andrés E. Caicedo, mookid, Carl MummertOct 14 '14 at 23:35

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, anomaly, Andrés E. Caicedo, mookid, Carl Mummert
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• I am sympathetic to those of you who don't know what is the "Archimedean principle" to which this user refers, but if you're going to thumb down, you could at least bother to ask "What do you mean 'Archimedean principle'?" You could even Google it. +1 to counteract laziness. – Kellen Myers Oct 13 '14 at 21:53

You might want to explain exactly what your textbook's / teacher's definition of "Archimedean principle" is. I will assume that it is this:

For any real number $x$ there is an $N$ such that $N>x$. See here.

If yours is slightly different (e.g. uses $\ge$) this should be fine. If it is significantly different, let me know and I will see if I can modify my answer.

To prove these two sets are equal (look at your equation -- it is an equation that says two sets are equal, right?), you need to really prove two things:

$$\displaystyle\bigcup_{n=2}^\infty [1/n,3-(2/n)] \subseteq (0,3)$$

and

$$(0,3) \subseteq \displaystyle\bigcup_{n=2}^\infty [1/n,3-(2/n)]$$

The first one is trivial, because $0<1/n$ and $3-(2/n)<3$ for all $n$, so the union of all those sets is contained in $(0,3)$. No element in any of the component sets $S_n=[1/n,3-(2/n)]$ is not also in $(0,3)$.

If you need to be formal, let $S_n=[1/n,3-(2/n)]$ and $S=\displaystyle\bigcup_{n=2}^\infty S_n$.

If $x\in S$, then for some $i$, $x\in S_i$. This means $0 < 1/i < x < 3-2/i < 3$. Thus $x\in(0,3)$.

The harder "direction" of the proof is the second $\subseteq$. So now take $x$ in $(0,3)$. We must prove it falls within that union somehow. This is where we apply the Archimedian principle.

First, take $N$ such that $N>1/x$. This means $1/N<x$. That helps us with the left hand side of our interval, the $1/n$ part.

It's a bit harder to deduce what to do about the other side, so let's break it down first:

We want $x<3-2/n$. We can rearrange that

$$x-3<-2/n$$

$$3-x>2/n$$

$$1/(3-x) < n/2$$

$$2/(3-x) < n$$

That means we need to take $N$ so that $N > 2/(3-x)$.

So we should take $N$ to not just be bigger that $1/x$, but really take:

$$N > \max\{1/x, 2/(3-x)\}$$

This is possible by the Archimedian principle. By our workings, this implies that $1/N<x<3-2/N$, which guarantees $x\in S_N$ and thus $x\in S$ (since $S=\cup S_n$).

That implies the second $\subseteq$ that we need to prove our set equality.

Note/edit: Sometimes the Archimedean principle (the $1/x$ or $1/\epsilon$ version) only applies to positive numbers, so it's important to note that for our $x$ (which is $0<x<3$), we know that $1/x$ and $1/(3-x)$ are both positive.