Metric $p := p(x,y)= \min(|x-y|, 1- |x-y|)$ $x,y \in [0,1)^2$. Prove metric space is compact.

Question

Help! I know that $X$ is Compact if every sequence in $X$ has a subsequence converging to a point in $X$. Also we have that $X$ is a bounded infinite subset in the real numbers.

I think it's quite a short proof, could anyone help please? Thank you!

Answer 1

Let $X=[0,1)$ with the given metric, and let $Y=[0,1]$ with the usual metric.

If $(x_n)$ is a sequence in X, then it has a subsequence $x_{n_k}$ which converges to $a \in Y$ since Y is compact.

1) If $a\ne1$, then $x_{n_k}$ converges to $a$ in X since

$\;\;\;|x_{n_k}-a|<\epsilon\implies d(x_{n_k},a)=\min\{|x_{n_k}-a|, 1-|x_{n_k}-a|\}<\epsilon$.

2) If $a=1$, then $x_{n_k}$ converges to 0 in X since

$\;\;\;|x_{n_k}-1|<\epsilon\implies 1-x_{n_k}<\epsilon\implies d(x_{n_k},0)=\min\{x_{n_k},1-x_{n_k}\}<\epsilon$.

Therefore $(x_n)$ has a convergent subsequence in X, so X is compact.