# How can one prove that $\bigcup^\infty_{n=1} [0, \frac n{n+1} ) = [0, 1)$?

Proving that if $$\omega \in [0, \frac{n}{n+1} ) \implies \omega \in [0, 1)$$ is easy but I'm struggling with the other way round. How do we know that $$\exists n: \omega < \frac{n}{n+1}$$? I think I should use the archimedean principle but not sure how?

• Do you know that $\displaystyle \lim_{n \to \infty} \frac{n}{n+1} = 1$? If so, use that.
– user169852
Commented Jun 13, 2021 at 20:25
• @Bungo How is this not an answer? Upvoting comments is not quite as satisfying as answers :) Commented Jun 13, 2021 at 20:28
• @SeverinSchraven It's a judgment call, but for me a one-line hint doesn't really qualify as an answer unless the OP asked for a one-line hint. Also, because there's no context, it's not clear if the OP knows about limits yet, which is why I posed the comment as a question.
– user169852
Commented Jun 13, 2021 at 20:29
• @Bungo I would disagree. If the one-line contains all the info needed, I would even prefer that. This really is the essence. But of course that is a matter of one's personal philosophy. Commented Jun 13, 2021 at 20:31
• @Bungo I do like the hint. Since we know that the limit of $\frac{n}{n+1}$ is 1, we can set $\epsilon = 1 - \omega$ and thus $\exists N$ such that $\forall n > N,$ $\omega < \frac{n}{n+1}$ Commented Jun 13, 2021 at 20:45

$$\omega < \frac{n}{n+1} \iff n \omega + \omega < n \iff \omega < n(1 - \omega) \iff \frac{\omega}{1-\omega} Take for instance $$n = \left\lfloor\frac{\omega}{1-\omega}\right\rfloor+1$$.
given $$\ x>0,\ y\in\mathbb{R},\ \exists n\in\mathbb{N}\cup\{0\}\$$ such that $$(n+1)x>y.$$
Choosing $$\ y=1\$$ and $$\ x = 1-\omega\ (>0),\$$ it follows that $$\ \exists\ n+1\in\mathbb{N}\$$ such that $$\ \frac{1}{n+1}<1-\omega,\$$ which implies that $$\ 1-\frac{1}{n+1} > 1 - (1-\omega) = \omega,\$$ i.e. $$\ \frac{n}{n+1}>\omega.$$