Why is $[0, 1] \cap \mathbb{Q}$ not compact in $\mathbb{Q}$?

Question

Statement: $[a, b] \cap \mathbb{Q}$ in $\mathbb{Q}$ is not compact. Thus the interior of all compact subsets of $\mathbb{Q}$ is $\emptyset$.

I am trying to understand the first sentence. I read that a closed subspace of a compact space is compact, so for example, consider the unit interval $[0, 1]$ which is a compact space. Take a closed subspace $[0, 1] \cap \mathbb{Q}$ of $[0, 1]$. This set is closed since it just consists of all the rational numbers in between $0$ and $1$, including $0$ and $1$. So it is a closed subspace of a compact space. But why isn't this compact?

Answer 1

Hint: Try finding a sequence in $\mathbb{Q}\cap[0,1]$ that does not have a convergent subsequence in $\mathbb{Q}\cap[0,1]$.

Answer 2

$\mathbb{Q}\cap[0,1]$ is dense in $[0,1]$

Answer 3

Hint: Find a sequence of closed, non-empty sets $C_n\subseteq [0, 1]\cap \Bbb Q$ for $n \in \Bbb N$ such that $C_n \subseteq C_{n-1}$ and $\bigcap_{n = 1}^\infty C_n = \emptyset$.

Answer 4

You can cover it with infinite number of open sets such that no open set overlaps (open intervals with irrational endpoints). If $\bigcup U_j = X$, but $U_j\cap U_k = \emptyset$ if $k\ne j$ you can't reduce the sets to a finite covering.

Answer 5

You said, "A closed subspace of a compact space is compact." and then you tried using $[0,1]$ as the compact space. But $[0,1]$ is compact as a subset of the reals, as a subset of $\mathbb{Q}$, it's what you are trying to understand. The countable covering of $[0,1]$ in $\mathbb{Q}$ by point sets in $[0,1] \cap \mathbb{Q}$ has no finite subcover in $\mathbb{Q}$ which covers the whole set, because none of the sets in the cover overlap.

Answer 6

The mistake in your argumentation is that the set $[0,1]\cap\mathbb Q$ is not closed in $[0,1]$. Rather, the closure of $[0,1]\cap\mathbb Q$ in $[0,1]$ is the complete interval $[0,1]$.

As non-closed set in a Hausdorff space, it cannot be compact.

Note that $[0,1]\cap\mathbb Q$ is closed in $\mathbb Q$, but then, $\mathbb Q$ is not a compact space, so again you don't get a closed subspace of a compact space.