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Is $[-1,1]\times [-1,1]$ a compact subset of $\mathbb{R}^{2}$ with respect to this distance? $$d((x,y),(a,b)) = \begin{cases} |x-a| & y = b \\ |x| + |y-b| + |a| & y \neq b \end{cases} $$

I have spent hours trying to make sense of this past problem with no luck. I have shown that $d$ is a distance. I was investigating if this subset is totally bounded. I got a hint to look at $B_{\frac{1}{10}}(1,1)$, which I determined to be $B_{\frac{1}{10}}(1,1) = \{(x,1) \in \mathbb{R}^{2} : x \in (\frac{9}{10},1]\} \subset [-1,1]\times[-1,1]$. However I do not see how this helps..

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    $\begingroup$ Look at the sequence $x_n=(1/2, 1/2+1/n)$. The distance between two of its points is $d(x_m,x_n)=1+\left|1/n-1/m\right|>1$. $\endgroup$
    – user517969
    Jan 2, 2018 at 20:36
  • $\begingroup$ Does that show that the space is not complete(and hence not compact)? I don't know what to do with such a sequence $\endgroup$ Jan 2, 2018 at 20:43
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    $\begingroup$ This sequence doesn't give any information about completeness, since it is not a Cauchy sequence. However, it is a sequence from which you cannot extract a convergent subsequence, since any two of its terms remain at distance larger than $1$. This shows that the space is not sequentially compact. $\endgroup$
    – user517969
    Jan 2, 2018 at 20:50
  • $\begingroup$ I understand, may I ask your thought process for finding such a sequence? $\endgroup$ Jan 2, 2018 at 20:53
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    $\begingroup$ (1) The second case in the definition of distance allows it to remain large because of that $|a|+|x|$. (2) exploit (1) by taking points that only fall in that case. $\endgroup$
    – user517969
    Jan 2, 2018 at 20:57

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