Is $[0,1] \cap \Bbb Q$ a compact subset of $\Bbb Q$?

Solution 1:

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 $[0,1]\setminus \{\sqrt{1/2}\}$.

Solution 2:

A set in a metric space is compact iff it's complete and totally bounded. $\Bbb{Q}\cap [0,1]$ is not complete.

Solution 3:

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 $\mathbb{Q}$.