# closed unit ball in $\mathbb{Q}$ is not compact

I want to prove that the closed unit ball $\overline{B_1(0)}=\{x\in\mathbb{Q}: |x|\le 1\}$ in $\mathbb{Q}$ is not compact ($\mathbb{Q}$ is endowed with the standard metric $d(x,y)=|x-y|$). To prove this, I need a sequence $(x_n)\subseteq \overline{B_1(0)}$ which has no subsequence $(x_{n_k})$ which converges in $\overline{B_1(0)}$.

I thought of something like $(x_n)\subseteq\overline{B_1(0)}$, $x_n\to \sqrt{2}$ and therefore $x_{n_k}\to \sqrt{2}$, but $\sqrt{2}\notin \overline{B_1(0)}$. Is it correct, how to define $x_n$? Do you have an idea of suitable sequences $(x_n)$ (maybe with other limit)?

• $\sqrt2>1$, so there cannot exist a sequence in $\overline{B_1(0)}$ converging to $\sqrt2$. However, you can use $1/\sqrt2$ and achieve the result. – Clayton Oct 26 '15 at 14:50
• There is no Cauchy sequence in the closed unit ball which has limit in $\mathbb R$ of $\sqrt 2$, as $\sqrt 2 - 1 > 0$. Try $L = 1/\sqrt 2$ where $a_n$ is the decimal expansion of $L$ to $n$ decimal places. $L$ is the only possibly limit of $(a_n)$ or any subsequence of $(a_n)$. But $L$ is not in the closed unit ball. – Simon S Oct 26 '15 at 14:50

## 1 Answer

You can't limit to a number outside of the interval $[-1,1]$. Instead try any irrational number $r$ such that $|r|<1$.

For example: Let $0.d_1d_2d_3\cdots = 1/\sqrt{2}$ (i.e. the decimal expansion of $1/\sqrt{2}$). Then $x_i = 0.d_1d_2\cdots d_i \in B_1(0) \subseteq \mathbb{Q}$. Clearly $x_n \to 1/\sqrt{2} \not\in\mathbb{Q}$.

Let $x_{n_k}$ be any convergent subsequence, then since $x_n$ converges (in $\mathbb{R}$) to $1/\sqrt{2}$ we must have that $x_{n_k} \to 1/\sqrt{2}$ as well. Thus there is no convergent subsequence converging to an element of the closed unit ball in $\mathbb{Q}$ (i.e. the closed ball isn't sequentially compact).