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So I'm having trouble starting the proof mainly because I don't know which proof technique to use. I thought about using the Principle of Mathematical Induction but it doesn't seem like the correct option. Any help is appreciated.

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What want to show two things:

(1) If $x$ is a number in the interval $(0,\infty)$ then it is in the set $\cup_{n=1}^\infty [1/n,n]$.

(2) If $x \in \cup_{n=1}^\infty [1/n,n]$ then $x \in (0,\infty)$.

The second one is easier. Take $x \in \cup_{n=1}^\infty [1/n,n]$. Then in particular there is some $N$ such that $x \in [1/N, N]$ because this is what set unions mean. Can you fill in the rest?

For showing that (1) is true, you need to use the archimedian principle. The principle states that for any number $x>0$ there is an $N$ such that $1/N < x$. Does this mean that $x \in [1/N, N]$?

(The above is slightly tricky. It does not necessarily follow, but can you see why? If $N$ doesn't work, you can find an $M$ that does work, however. This is related to splitting up between cases $0<x<1$ and $x \geq 1$.)

To finish (1), if we have an $N$ such that $x \in (0,\infty)$ means that $x \in [1/N,N]$ can you see why $x \in \cup_{n=1}^\infty [1/n, n]$?

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  • $\begingroup$ We never learned Archimedes Principle but it all seems to make sense. I'll have to review it later. Thank you very much! $\endgroup$ – Robin Oct 6 '15 at 20:28
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Hint: Condition on whether $x \in (0, \infty)$ is less than or equal to or greater than $1$, and then use the Archimedes principle.

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When you have an equality of sets you almost always have to prove the two inclusions $\subseteq$ and $\supseteq$.

We have $[\frac{1}{n},n]\subseteq (0,+\infty)$ for all $n$ which implies $\bigcup [\frac{1}{n},n]\subseteq (0,+\infty)$.

Now for the other inclusion take $x\in (0,+\infty)$. If $x\geq 1$ there is a $n$ such that $n>x$ and $x\in [\frac{1}{n},n]$. If $x<1$ then $\frac{1}{x}>1$ and you can use the same reasoning.

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