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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?

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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.

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$]-1,2[$ is a finite cover, but it is not among your $U_n$. You are arguing that your $U_n$ form a cover, which is correct in that every rational in $]-1,2[$ is in at least one of the $U_n$. A finite subcover would be a collection of a finite number of your $U_n$ which covered all the rationals. As your $U_n$ are nested, the union of a finite number of them is just the one with the highest index. Any finite collection will have one with the highest index, so you need to show that there is at least one rational not in that set. That proves that this finite collection is not a cover.

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Your idea is correct. Simply note that your sequence is decreasing and for each $n$ there is a rational number in the interval $(\sqrt{\frac{1}{2}}-\frac{1}{n}, \sqrt{\frac{1}{2}}+\frac{1}{n})$.

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More generally (and easily), consider any irrational number $\alpha\in[0,1]$. There is a decreasing sequence $(b_n)$ in $\mathbb{R}_{>0}$ (even in $\mathbb{Q}$, actually) such that $\alpha+b_0<1$ and $\lim_{n\to\infty}b_n=0$.

Then you can consider the open sets $$ U_n=(-1,\alpha)\cup(\alpha+b_n,2) $$ Clearly, $U_n\subseteq U_{n+1}$, for all $n$, and so $$ U_{n_1}\cup U_{n_2}\cup \dots\cup U_{n_k}=U_m $$ where $m=\max\{n_1,n_2,\dots n_k\}$. Since $\alpha<\alpha+b_m$, there is a rational number $q\in\mathbb{Q}\cap[0,1]$ with $\alpha<q<\alpha+b_m$. Therefore $\{U_{n_1}, U_{n_2},\dots, U_{n_k}\}$ cannot be a cover of $\mathbb{Q}\cap[0,1]$.

On the other hand, $\{U_n\}$ is an open cover for $\mathbb{Q}\cap[0,1]$. Indeed, let $q\in\mathbb{Q}\cap[0,1]$. If $q<\alpha$, $q$ belongs to every $U_n$. If $q>\alpha$, there exists $n$ such that $q>\alpha+b_n$, so $q\in U_n$.

Note that compactness means every open cover has a finite subcover; the finite covers do, but this is irrelevant.

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  • $\begingroup$ Thanks for your more general solution! $\endgroup$ – Staki42 Nov 28 '16 at 15:37

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