Is the following metric space compact?

Question

$M = [-1,1]\times[-1,1], d((x,y),(a,b))=\begin{cases}|x-a|& y=b\\ |x|+|a|+|y-b| & y\not= b\end{cases}$

I feel like this is not compact and have being trying to show this by showing it's not sequentially compact. If we take a sequence $(1,1)\to(1/2,1/2)$ along a straight line, then we can't find $N\in \mathbb N : d(a_n,(1/2,1/2))<\epsilon \forall n>N$ but I am confused here

Is there an easy way to see that $d(\cdot,\cdot)$ is a metric? — kimchi lover, Jan 01 '18 at 20:25

Shashi · Accepted Answer · 2018-01-01T20:33:45.130

3

You are on the right track. However your sequence is a little bit vague.

Take $x_n=(1, \frac{1} {n} )\in M$. You can easily see that for $n\neq m$ we have $$d(x_n, x_m) = 2+\bigg |\frac{1}{n}-\frac{1}{m}\bigg |\geq 2$$ So any subsequence you can think of cannot be Cauchy. Since convergent sequences are Cauchy, being not Cauchy implies not convergent. So you can never find a convergent subsequence. So $M$ is not sequentially compact. In metric spaces sequentially compact and compact is equivalent, so $M$ is not compact as well.

edited Jan 01 '18 at 20:33

answered Jan 01 '18 at 20:26

Shashi

8,738

Thank you I was mainly confused about whether my idea was how one would show that something isn't sequentially compact – Jan 01 '18 at 20:30
@Spudnick it is, but one needs to be a little bit careful since you want that no subsequence to be Cauchy. I added more explanation to help you understand better. – Shashi Jan 01 '18 at 20:34

Is the following metric space compact?

1 Answers1