Can I show these questions (is a set open or closed WRT metric) a faster way? I have the metric:
$$d((x,y),(a,b))=|y-b|\text{ if }a=x\text{ else }|y|+|b|+|x-a|$$
I have been asked the following questions:


*

*Is the set $\{0\}\times(0,1)$ open with respect to this metric? Is it closed?

*Is the set $\{0\}\times(-1,1)$ open WRT this metric? Is it closed?

*Is the unit ball $\{(x,y)\in\mathbb{R}^2|x^2+y^2<1\}$ open in this metric


For the third one I am allowed to use (given) $d((x,y),(a,b))\ge||(x,y)-(a,b)||$ where $||\bigdot||$ is the Euclidean Norm (Pythag)
For part 1, I spent A LONG TIME (more than I want to admit) showing:
$\forall a\in M\exists r_m>0:y\in B(x,r_y)\implies y\in M$
Where $M=\{0\}\times(0,1)$
(That is to say that $M$ is a neighborhood of all its points)
There is a mark scheme for this, bt it's not very informative it gives (for the first part)
${0}\times(0,1)=B((0,\frac{1}{2}),\frac{1}{2})$
Which I'll be honest, I still cant see how one would spot that. Additionally the metric is quite tedious to draw, without spending time investigating it (which I did in the end, longer than I would like) and studying when open balls are vertical lines and when they are almost rotated squares, less the top vertex + a point above the square (which I didn't investigate WRT when this happens) I can't see how you'd spot that.
It also says "it is not closed because $(0,0)$ is no in any ball $B((0,\frac{1}{2}),r)$ with $r>\frac{1}{2}$" I'm not sure where or why this is, I have just realised that the sadly, after spending much more than one hour (I don't want to confess how long...) I have actually only done half of the first question.
So what have I missed? I've been drawing some pictures, but not useful ones (if they were useful I'd not be writing this)
 A: Well, one may not spot the fact that $M$ is just a single open ball, but going the natural way (i.e. trying to show that there is a ball around each point), you should check what is the distance of a point $(u,v)$ not in $M$ from a given point $(0,y)\in M$? Two cases arise: Either $u\ne 0$; then $d((0,y),(u,v))=|y|+|v|+|a|\ge |y|$. Or $u=0$; then $y\ge 1$ or $y\le 0$, hence $d((0,y),(u,v))=|v-y|\ge \min\{|1-y|,|y|\}$. We conclude that $B((0,y),r)\subseteq M$ where $r=\min\{y,1-y\}$. And now you may stumble upon the symmetry and see that in fact $M=B((0,\frac12),\frac12)$ (but this is not needed, a different ball for each $y$ would still be fine).
$M$ is closed iff its complement is open. That would mean especially, that some open ball around $(0,0)$ is disjoint from $M$. But such a ball $B((0,0),r)$ with $r>0$ (and wlog. $r<1$) would contain $(0,\frac r2)$ because $d((0,0),(0,\frac r2))=\frac r2<r$. You could do the same with $(0,1)$. Even with a metric as weird as this, it should have occured that the first candidates to check if they are boundary points are $(0,0)$ and $(0,1)$.
For the second (and third) question note that the balls around a point $(x,0)$ look different from those around $(x,y)$ with $y\ne 0$: For small enough radii, the latter are just vertical line segments. But for $(x,0)$ even a small $r$-ball contains also points $(u,v)$ with $u\ne x$!
