Open cover of $\mathbb{Q} \cap [0,1]$

Question

Let $\alpha \in [0,1]$ such that $\alpha$ is irrational. Then an open cover of $\mathbb{Q} \cap [0,1]$ is $\{-1,\alpha-\frac{1}{n} \big | n \in \mathbb{N} \} \cup (\alpha,2)$. I found this solution and am a bit skeptical of it. How can we be sure that $(-1,\alpha-\epsilon) \cup (\alpha, 2)$ has no rationals in between?

Answer 1

You are right, that $(-1, \alpha-\epsilon)\cup (\alpha, 2)$ may not cover all of $\Bbb Q\cap[0,1]$, for the reason you said. But the construction says nothing about $\epsilon$, which you made up and inserted into the construction yourself.

The covering includes an interval $\left(-1, \alpha-\frac1n\right)$ for each integer $n$. Let us consider an irrational number $r\lt \alpha$ which you think might not be covered. Then let $x = \alpha-r$, so $x>0$. Then by the Archimedean property of the real numbers, there is an integer $N$ that is greater than $\frac1x$, so $\frac1N < x$ and $\alpha - \frac1N > \alpha - x$. Then the interval $\left(-1, \alpha - \frac1N\right)$ contains $r$, because $\alpha - \frac1N > \alpha - x = r$. So for each irrational $r < \alpha$, $r$ is covered by some element of the covering.

Perhaps what you are trying to do is invent an “infinitesimal” $\epsilon$ so that $\frac1{\epsilon}$ is less than $\frac 1N$ for every positive integer $N$, but there is no such $\epsilon$ in the real numbers.