Show $\forall \delta > 0, \exists n \in \mathbb{N}$ such that $\frac{1}{n} < \delta$ The question is in the title, but I have no idea how to solve it, so a few hints would be appreciated, thanks.
Suppose the statement is false. Then we have some $\delta&gt;0$ such that $\delta^{-1}&gt;n$ for all $n$. This contradicts the Archimedean property. Assuming $\mathbb{R}$ is defined axiomatically, this property is easily derived as a consequence of the least upper bound axiom.
Given $\delta$, let $k = \frac{1}{\delta}$

$$\delta = \frac{1}{\frac{1}{\delta}} = \frac{1}{k}$$

Consider $ \frac{1}{\lfloor k+1 \rfloor}$.