I have to prove that $\mathbb Q \cap [0,1]$ is not compact, directly with the definition with open covers (I am not allowed to use theorems like Heine-Borel).
My attempt:
So I need to find an open cover of $\mathbb Q \cap [0,1]$ that has no finite subcover. I assume that we need to this by something like approximating an irrational number, i.e.
$$
U_n := \left]-1, \frac{\sqrt2}{2} – \frac{1}{n}\right[ \cup \left]\frac{\sqrt2}{2}+\frac{1}{n},2\right[
$$
This is a open cover of $\mathbb Q \cap [0,1]$ but I am not sure if this does not have a finite sub cover, like for example just $\mathopen]-1,2\mathclose[$
Any hints?
Best Answer
Your answer is correct. Suppose that $U_{n_1}, \dots, U_{n_p}$ is a finite subcover with $n_1 < \dots < n_p$. Then $$U_{n_1} \cup \dots \cup U_{n_p} = ]-1, \frac{\sqrt2}{2} - \frac{1}{n_p}[ \cup ]\frac{\sqrt2}{2}+\frac{1}{n_p},2[$$ and a rational between $\frac{\sqrt2}{2} - \frac{1}{n_p}$ and $\frac{\sqrt2}{2}$ won't be in the subcover.