1
$\begingroup$

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)?

$\endgroup$
2
  • 2
    $\begingroup$ $\sqrt2>1$, so there cannot exist a sequence in $\overline{B_1(0)}$ converging to $\sqrt2$. However, you can use $1/\sqrt2$ and achieve the result. $\endgroup$ – Clayton Oct 26 '15 at 14:50
  • $\begingroup$ 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. $\endgroup$ – Simon S Oct 26 '15 at 14:50
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.