How do I visualize a set in metric space?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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 $f_2(x) = x+1$?