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If I'm given a metric, say the discrete metric $\text d_0(x,y):=\begin{cases} 0, & \text{if }\vec x= \vec y\\ 1, & \text{if } \vec x \neq \vec y\ \end{cases}$ and want to visualize a set given by $\mathbb{S}^{1}_0(\vec0) = \lbrace\vec x \in \mathbb{R}^2 |\text d_0(\vec x, \vec 0) = 1\rbrace$, how can I grasp the of definition of the set? Since I can't even begin to imagine it, any hint would be highly appreciated.

psyph
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    It is all points except the origin? Set $y=0$ in your formula and figure out what $x$ satisfy the definition. There are no general tricks, just doing the grind. – copper.hat Jun 11 '19 at 16:08

4 Answers4

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For most metrics you will encounter on ${\mathbb R}^2$ you can sketch sets fairly easily. Start off by working out which points must lie on the boundary of the set you want to sketch and work from there, checking if there are 'holes' in the interior somehow.

For the example you give, since every point except $0$ is a distance $1$ from $0$ the 'sphere' in this case is the whole of ${\mathbb R}^2$. You can sketch that as the usual plane for ${\mathbb R}^2$, with the centre point $0$ highlighted to indicate its exclusion.

You might also want to look at the shapes of the unit ball under various metrics and how they move from diamond-shape to circular to square-shaped.

postmortes
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    So I'm just looking for all $\vec x$ that satisfy the condition that the distance between them and $\vec y$ with $\vec y = \vec 0$ in my case is equal to 1, right? Now I see, that statement will be true for all $x \in \mathbb{R}^2 \ \lbrace 0 \rbrace$ – psyph Jun 11 '19 at 16:30
  • Also: thanks so much for your answer. – psyph Jun 11 '19 at 16:30
  • @M.P. yes, you've got it :) – postmortes Jun 11 '19 at 17:10
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In order to comprehend what such a set would "look" like (I used the quotes at look, because in many-dimensional or infinite dimensional spaces you can only think of an essential structure), one has to grind the defined metric and the definition of the set and understand what it means.

Specifically, you have that $d_0(x,y) = 1, \; x \neq y$. So, for $x \in \mathbb R^2$ to be $d_0(x,0) = 1$, it should be $x \neq 0$. Thus, all non zero elements of $\mathbb R^2$ belong in the set $\mathbb S^1_0(0)$.

I skipped the vector arrows for the sake of simplification of notation.

Rebellos
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Fix $x=0$.

If $d_0(0,y)=1$, then $y\neq 0$ by the definition of $d_0$.

Conversely, $d_0(0, 0) = 0\neq 1$.

Thus $\mathbb S_0^1 = \mathbb R^2\setminus \{0\}$.

klirk
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Thanks so much for all the answers, which were all very helpful. Just to make sure I understood: for $\text{d}_0(\vec x,\vec y): = |x_1 - x_2|+|y_1 - y_2|$, given I'm looking for the same set, do I just have two parallel linear lines with $f_1(x)=x-1$ and $f_2(x) = x+1$?

psyph
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    You should ask that as a different question -- it's not an answer and the review queues will pick it up and close/delete it. However, this $d_0$ is not a metric, so can you check if you wrote it correctly (hint: $d_0((a,a),(b,b))=0$ for any $a$ and $b$) – postmortes Jun 11 '19 at 17:13
  • Thanks for your advice since I'm very new here. Can I transfer my second question into the comment section, then? – psyph Jun 11 '19 at 17:17
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    You can, but since it's a question in its own right you should ask it separately. That said, you should also search the site, as I think you'll find it's already been asked and there's probably a nice answer with some graphs to show you what's going on. This question: https://math.stackexchange.com/questions/520002/unit-ball-with-p-norm might help you – postmortes Jun 11 '19 at 17:19