Question about the radius of an open ball that is less than "zero" in Euclidean metric space.

Question

define a metric:

$d(\mathbb{x}, \mathbb{y})=|\mathbb{x}|+|\mathbb{y}|$, where $\mathbb{x}, \mathbb{y} \in \mathbb{R}^k$ when $\mathbb{x} \neq \mathbb{y}$ and $d(\mathbb{x},\mathbb{x})=0$.

consider the open ball in $(\mathbb{R}^k, d)$, we say

$B_{r}(\mathbb{x}) = \{\mathbb{y}\in \mathbb{R}^k : |\mathbb{x}|+|\mathbb{y}|<r\}$ where $r>0$.

if $r<|\mathbb{x}|$, we get:

$B_{r}(\mathbb{x}) = \{\mathbb{y}\in \mathbb{R}^k : |\mathbb{y}|<r - |\mathbb{x}|<0\}$

the question is which of the following is true? (1) or (2):
(1) if $r<|\mathbb{x}|$, then $B_{r}(\mathbb{x}) = \{\mathbb{x}\}$.
(2) if $r<|\mathbb{x}|$, then $B_{r}(\mathbb{x}) = \{\varnothing \}$.

and explain why.

Answer 1

if (1) is true:
given $p \in \{\mathbb{x}\}$,
==>then make $0<\epsilon<r$,
==>$N_{\varepsilon}(p) \in \{\mathbb{x}\}$
$p$ is an interior point of $\{\mathbb{x}\}$, which implies $B_{r}(\mathbb{x})$ is open

if (2) is true: $\{\varnothing \}$ is open ,which implies $B_{r}(\mathbb{x})$ is open

for both (1) and (2) can get an open set, so (1) and (2) is true, and $\{\varnothing\} \subset \{\mathbb{x}\}$

we get:

if $r<|\mathbb{x}|$, then $B_{r}(\mathbb{x}) = \{\mathbb{x}\}$.

my answer is deduced from the consequence to the reason.

i don't know if there exists a definition or a theroem to show this is true or false.

Answer 2

$d(\mathbb{x}, \mathbb{y})=|\mathbb{x}|+|\mathbb{y}|$ does not define a metric on $ \mathbb R^k$ !

Reason: $d(\mathbb{x}, \mathbb{x})=2|\mathbb{x}| \ne 0$ if $\mathbb{x} \ne \mathbb{0}$.

Answer 3

An intuition for this metric is an "all roads lead to Rome" kind of star graph centered at zero. If you want to go from $x$ to $y\ne x$, you have to walk to $0$ first and then to $y$, for a total distance of $|x|+|y|$.

If you don't have enough energy to even walk to zero (i.e. $r<|x|$), then you can't make it anywhere except for $x$ itself. So $B_r(x)=\{x\}$.

Again, formally: Let $0<r<|x|$. Clearly $x\in B_r(x)$ since $d(x,x)=0$. But if $y\ne x$ then $d(x,y)=|x|+|y|\ge |x|>r$, so $y\notin B_r(x)$. Thus $B_r(x)=\{x\}$.

Side note: $B_r(x)=\{\emptyset\}$ is not type correct, since $\emptyset$ is not a point in $\Bbb R$. Other reasonable choices for $B_r$ include $\{0\}$ and $\emptyset$, but $\{\emptyset\}$ could not possibly be correct.