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The title said it all. I have come up with a solution, but I cannot figure out some details. Please help me out and comment on my solution. Feel free to leave your own solution so that I can also learn from you.

Show that $[0,1] \cap \mathbb{Q}$ is not compact in $[0,1]$.

Solution: Let $S =[0,1] \cap \mathbb{Q}$. Note that $S$ is countable.(Is it? If yes, how to prove it rigorously? If not, is the remaining part of my solution still true?) Then we write $S=\{ r_n : n \in \mathbb{N} \}$. We now construct an open cover that does not have a finite subcover. Consider $r_n \in S$, where $r_n \neq 0, 1$, then we define $p = \min \left\{\frac{d(r_{n-1},r_n)}{2}, \frac{d(r_n, r_{n+1})}{2}\right\}$. Then consider $$\left(\bigcup_{r_n \in S, r_n \neq0,1}B(r_n,p)\right)\cup 0\cup 1$$ is an open cover of $S$. But there exists no finite subcover. Thus, $S$ is not compact in $[0,1]$.

Apart from the above question, if I change the metric space from $[0,1]$ to $\mathbb{R}$, can I still make the same conclusion? This confuses me as I did not really consider the metric space in my solution!

Thanks in advance.

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    $\begingroup$ $\Bbb Q$ is countable, and thus, so is any (infinite) subset of $\Bbb Q$. In particular $S$. $\endgroup$ – Arthur Oct 7 '14 at 12:57
  • $\begingroup$ $\ldots \cup 0\cup 1$ can't be right, because $0$ and $1$ are not sets and you cannot take the union of them. Normally one might suppose that you meant $\{0\}$ and $\{1\}$, but that doesn't work here because you are constructing an open cover and $\{0\}$ and $\{1\}$ are not open sets. $\endgroup$ – MJD Oct 7 '14 at 13:44
  • $\begingroup$ @MJD: Noted. So is there any way to modify the above open cover so that the remaining is still true? $\endgroup$ – Nighty Oct 7 '14 at 13:47
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    $\begingroup$ Yes, it's unreasonable. You can enumerate the rationals, so that for each element of $[0,1]\cap \Bbb Q$ you have exactly one $r_k$. But you cannot enumerate the rationals in order so that $r_k < r_j$ when $k<j$, because there is no ‘next’ rational after a given rational. (See the first paragaph of this for an explanation). $\endgroup$ – MJD Oct 7 '14 at 14:21
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    $\begingroup$ Also your enumeration $r_k$ can have a beginning $(k=0)$ but it cannot have an end, because if it did you would have a finite enumeration, but the set you are enumerating is infinite. $\endgroup$ – MJD Oct 7 '14 at 14:25
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Here's an easy solution:

HINT: Compact sets in metric spaces are closed.

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  • $\begingroup$ Let $X$ be the metric space we are considering. $X\setminus S$ should contain all the irrational numbers in $[0,1]$. But any open ball of $x \in X\setminus S$ is not contained in $X\setminus S$. $X\setminus S$ is not open $\Rightarrow S$ is not close? $\endgroup$ – Nighty Oct 7 '14 at 13:40
  • $\begingroup$ Yes, if the complement is not open, the set is not closed. $\endgroup$ – Asaf Karagila Oct 7 '14 at 13:50
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$x_n ={\lfloor\sqrt{2}n\rfloor\over 2n}\in[0,1]\cap\mathbb{Q}\forall n\in\mathbb{N}$, $x_n\to {1\over\sqrt{2}}\notin [0,1]\cap\mathbb{Q}$

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  • $\begingroup$ why sequence Xn converges to 1/root 2? $\endgroup$ – jessie Oct 28 '15 at 1:28
  • $\begingroup$ @jessie, $\lfloor \sqrt{2} n \rfloor = \sqrt{2} n - r $ where $0 \leq r < 1$. When $n$ is big, you can throw away $r$, then you get the limit. You can also follow the $\varepsilon - \delta$ definition of a limit. $\endgroup$ – swoopin_swallow Apr 12 '17 at 2:02

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