Consider the family of metric function $d: \mathbb{R}^d \times \mathbb{R}^d \rightarrow [0, 1] $ with the following properties:
- Coincidence: $d(x, y) = 0 \Longleftrightarrow x = y, \forall x, y$
- Symmetry: $d(x, y) = d(y, x), \forall x, y$
- Triangle inequality: $d(x, z) \leq d(x, y) + d(y, z) , \forall x, y, z$
For a fixed set of vectors $\left\{ \textbf{v}_1, ..., \textbf{v}_n \right\}$ where $\mathbf{v}_i \in \mathbb{R}^d$ and no two vectors are the same ($\forall i, j: \mathbf{v}_i \neq \mathbf{v}_j$), prove that there are infinite many points in $\mathbf{v}_i$'s neighborhood. That is, there are infinite points in $\mathbb{R}^d$ that satisfy the following, for a fixed $i$: $$ d(\textbf{x}, \textbf{v}_i) < d(\textbf{x}, \textbf{v}_j), \forall i \neq j $$
If this is not true, what additional properties does $d(., .)$ metric need to have for this to be true?
For example, if $d$ is a dot product, this is definitely true since, for a fixed $i$, there are infinite $\mathbf{x} \in \mathbb{R}^d$ that satisfy the following: $$ \textbf{x}.\textbf{v}_i < \textbf{x}.\textbf{v}_j, \forall i \neq j $$