Let $\mathbb{R}^{2}$ with British Rail metric. Show that the unit closed ball $B_{\leq 1}(\overline{0})$ is not compact.

Question

Let $X=\mathbb{R}^{2}$ and $d$ a metric given by $$ d(\overline{x},\overline{y}):=\left\{ \begin{array}{ll} \left\|\overline{x}\right\|+\left\|\overline{y}\right\| & \mbox{if }\overline{x}\neq \overline{y} \\ 0 & \mbox{if }\overline{x}= \overline{y} \end{array}\right. $$ Show that the unit closed ball $B_{\leq 1}(\overline{0})$ is not compact.

Answer 1

HINT: Let $D=\{x\in X:\|x\|=1\}$. Show that $D$ is an infinite, closed, discrete subset of $B_{\le 1}(\overline 0)$, and explain why this shows that $B_{\le 1}(\overline 0)$ is not compact. If you get completely stuck, there’s a further hint in the spoiler-protected block below.

Extra HINT: What is $d(\overline x,\overline y)$ if $\overline x$ and $\overline y$ are distinct points of $D$?

Answer 2

Let the sequence $x_n=(1-\frac{1}{n},0)$ defined for $n \geq 1$. If $B_{\leq 1}(\overline{0})$ is compact, we can find $\phi$ strictly increasing such that $x_{\phi(n)} \mapsto y$. But $d(x_{\phi(n)},x_{\phi(k)})\geq \frac{1}{2}$ if $n \neq k$, so it's impossible.