Is $[0,1] \cap \Bbb Q$ a compact subset of $\Bbb Q$? Is $[0,1] \cap \Bbb Q$ a compact subset of $\Bbb Q$?  I get the feeling it isn't compact, but I can't figure out a way to prove this.  I understand that,  in $\Bbb Q$, any open set $U = \Bbb Q \cap O_i$ such that $O_i \subseteq R$.  But I can't figure out a way to construct a finite set that doesn't contain the intersection set.
 A: Let $r$ be an irrational number in $[0,1]$. There exists $N \in \mathbb{N}$ such that
$$ \frac{1}{N} \leq \min\{r, 1-r\} = \text{dist}\big(r, \mathbb{R}\backslash [0,1]\big).$$
The sets 
$$U_n := \left(-1, r-\frac{1}{n}\right) \cup \left( r+ \frac{1}{n}, 2\right),~~~~~n \geq N$$ are open. The collection of $U_n$ forms an open covering of $[0,1] \cap \mathbb{Q}$ with no finite sub-covering. Hence $[0,1] \cap \mathbb{Q}$ is not compact in $\mathbb{Q}$.
A: A set is sequentially compact if every sequence contained in the set have a subsequence that converges to a point of the set. 
Because the set proposed is only composed of rational numbers then any sequence that converges to some irrational number doesn't have any subsequence that converges to a point of the set (because for convergent sequences every subsequence converges to the same point).
A: Compact spaces have the property that they are closed in any Hausdorff superspace. Since ${\bf R}$ is Hausdorff and $[0,1]\cap {\bf Q}$ is not closed in ${\bf R}$, it cannot be compact.
To see that directly, you do the same as you would with, say $[0,1]\setminus \{\sqrt{1/2}\}$.
A: A set in a metric space is compact iff it's complete and totally bounded. $\Bbb{Q}\cap [0,1]$ is not complete. 
A: Well, let's figure it out knowing it can't be true.
In R a set is compact iff it is closed and bounded.  [0,1] restricted to Q is bounded but it isn't closed.  (Which is enough... we are done.  But ...yeah, I know, it doesn't feel we are done.  It'd feel better if we could find an open cover that has no finite cover.)
Well, let's find a limit point of [0,1] restricted that isn't in the set.  Well, that'd be any irrational $x; 0 <x <1$.  Let's make an open cover that "tends" toward x in a "limit pointy sort of way".
Oh, all right ... I'll out and say it.  Let's use the 1/n to get open sets that come within 1/n of x.  As x is a limit point, these will get arbitrarily close to x and cover all but x.  But as x isn't in the set it will cover the set.
Okay, let's formalize that.
Let $U_n = (-\infty, x -1/n)\cup (x+1/n,\infty) $.  This collection $\{U_n\} $ is an open cover as it covers everything but $x$ which wasn't in the set to begin with.  Any finite subcollection, $ \{U_{n_i}\} $ will have an $m=\max (n_i) $ and so $ [x-1/m,x+1/m]\cap [0,1]\cap \mathbb Q \subset [0,1]\cap \mathbb Q$ will not be covered.
So that's an open cover with no finite subcover.
A: This is a particular case of a more general affirmation, namely: Every compact set in $\mathbb{Q}$ has an empty interior. 
To prove this use the fact that in any open interval $(a,b)\subseteq\mathbb{R}$ there is a sequence of rationals converging to an irrational (which will of course contain no subsequence converging to a rational), and recall that in any compact space every sequence contains a convergent subsquence.
