Why aren't the rationals a compact subset of $\mathbb{R}$?

Question

We define a compact subset of some normed vector space $V$ to be any subset $S$ where every sequence $\{\mathbf{x}_{n}\}$ in $S$ has a subsequence which converges to some $\mathbf{x}$ in $S$.

Then it seems like it must be a duplicate but I can't find a good reason anywhere, why aren't the rationals a compact subset of $\mathbb{R}$? How can I most easily show this? I'm trying to better understand the notion of compact subsets.

Answer 1

A compact space is complete.

Another reason: a compact subspace is closed. And precisely, the closure of $\mathbf Q$ is $\mathbf R$.

Answer 2

Another reason: a compact subset of the reals is closed and bounded (Heine-Borel theorem), and the rationals are distinctly not bounded.

Answer 3

A space must be complete in order to be compact. Complete means that every Cauchy sequence converges, that is every sequence "approaches" to a point that belongs to the considered set. For example, the sequence $$ a_n=\bigg(1+\frac{1}{n} \bigg)^n $$ is a sequence in $\mathbb{Q}$ but it does not converge into $\mathbb{Q}$. Therefore, $\mathbb{Q}$ is not complete and thus is not compact.