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Is $[-1,1] \times[-1,1]\subset\Bbb R^2$ a compact subset with regards to the distance function

$$d\bigl((x,y),(a,b)\bigr):=\begin{cases}|x-a| & \text{if }y=b\\|x|+|y-b|+|a| & \text{if }y\neq b\end{cases}$$.

Well the first thing I would note is that the subset is closed and then maybe try to show that for both cases the distance is bounded above

$|-1-1|=2$ furthest distance if $y=b$

$1+2+1=4$ furthest distance if $y\neq b$

So it's bounded above by four, and so closed and bounded and then by the Heine-Borel Theorem also compact? Is this close to being correct?

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Hint

For each $a \in [-1,1]$ the set $(-1,-1) \times \{ a\}$ is open.

Show that for some small $r >0$ $$ \left( [-1,1] \times [-1, -1+r) \cup [-1,1] \times (1-r, 1] \right)\cup \left(\bigcup_{a \in [-1,1]} I_a \right) $$ is an open cover which doesn't have a finite subcover.

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Hint: If you want to use the Heine-Borel characterisation: this space is not totally bounded. You cannot cover it with finitely many balls of diameter less than $1$, for instance.

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