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We know that $[0,1]$ is compact in $\mathbb{R}$, then for an arbitrary open cover, $O_i,\; i \in \{1,2,...\}$, there exists a finite subcover, $O_{i_k},\; k=\{1,2,...,n\}$ such that $[0,1] \subset \bigcup_{k=1}^nO_{i_k}$. Since for each $Q_{i_k}$ is open in $\mathbb{R}$, $O_{i_k}\cap \mathbb{Q}$ is open in $\mathbb{Q}$. Now, $$\begin{align*} [0,1] \subset \bigcup_{k=1}^nO_{i_k} \Rightarrow [0,1]\cap \mathbb{Q} &\subset \left(\bigcup_{k=1}^nO_{i_k}\right)\cap \mathbb{Q} \\ & = \bigcup^n_{k=1}(O_{i_k}\cap\mathbb{Q}) \end{align*}$$ Since $O_{i_k}\cap\mathbb{Q}$ is open in $\mathbb{Q}$ and we have a finite union of them, $ [0,1]\cap \mathbb{Q} $ is compact in $\mathbb{Q}$.

But Clearly, it is not compact when I checked the answer, meaning that I made a mistake somewhere. Can anyone please fix it?

Thanks

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Hint: $$ [0, 1]\cap \Bbb Q = \big([0, \frac1\pi) \cup (\frac1\pi, 1]\big)\cap \Bbb Q $$ where $[0, 1/\pi) \cup (1/\pi, 1]$ clearly isn't compact.

Your mistake is to assume that a collection of open sets in $\Bbb R$ that covers $[0,1]\cap \Bbb Q$ must cover $[0,1]$.

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  • $\begingroup$ Nice. +1 for only pointing out the error in OP's solution and not giving him your own solution. $\endgroup$ – 5xum Apr 4 '16 at 8:45
  • $\begingroup$ @Arther But I started the proof from defining open coverings that cover $[0,1]$ and they covers $[0,1]\cap\mathbb{Q}$...? $\endgroup$ – user1292919 Apr 4 '16 at 8:52
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    $\begingroup$ @user1292919 But that means you are not considering all possible open covers of $[0,1]\cap \Bbb Q$; you are only considering the ones that happen to cover the entire $[0,1]$. To show that a set is compact, you need to show something about all open covers, not just the nice ones. $\endgroup$ – Arthur Apr 4 '16 at 8:53
  • $\begingroup$ @Arthur Oh! I understood now Thanks :) $\endgroup$ – user1292919 Apr 5 '16 at 0:27

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