There is a natural number n with 1/n below any positive epsilon

Statement

Let $\varepsilon$ be a real number with $\varepsilon > 0$. Then there is a natural number $n \ge 1$ with $\dfrac{1}{n} < \varepsilon$.

Remarks

Proof: since the reals are Archimedean, there is a natural number $n > 1/\varepsilon$. Then $n \ge 1$ and $1/n < \varepsilon$. Thus $\inf \{1/n : n \ge 1\} = 0$. There are no positive infinitely small real numbers. The same holds in each Archimedean ordered field.

Explanations

L1 human

The fractions $1, 1/2, 1/3, \dots$ get as small as you want. For $\varepsilon = 0.001$, the number $n = 1001$ works, because $1/1001 < 0.001$.