Investigations into infinite metric bases (Part 1/2) | The Hitchhiker's Guide to Computable Mathematics

Let $(X,d)$ be a metric space. Let $A \subseteq X$ be such that for all $x,y \in X$ the following implication holds:

$\left(\forall a \in A \ d(x,a) = d(a,y) \right) \implies x=y$

then we say $A$ is a metric base for $(X,d)$ . A sequence $\alpha$ in $X$ is called a dense sequence iff $\mathrm{Im}\ \alpha$ is a dense set. A metric base $A$ is a finite metric base iff $A$ is a finite set. Finding a finite metric base (if it exists) in a general metric space is already an interesting challenge and this will be explored in more detail in another post. Some information about how finite metric bases relate to computability can be found in another post. The question of existence of infinite metric bases however seems a bit less challenging due to the following easy result.

Proposition: Let $(X,d)$ be a metric space. Let $\alpha$ be a dense sequence in $X$ . Then $\mathrm{Im}\ \alpha$ is a metric base for $(X,d)$ .

Proof. Let $\alpha$ be a dense sequence and set $A = \mathrm{Im}\ \alpha$ . Let $x,y\in X$ . Suppose $d(x,a) = d(y,a)$ for all $a \in A$ . Since $\alpha$ is dense, there exists a sequence $(\beta_i)$ in $A$ such that $\lim_i \beta_i = x$ . This is equivalent to $\lim_i d(\beta_i, x) =0$ . Since $d(x,a) = d(y,a)$ for all $a \in A$ we have $\lim_i d(\beta_i, y) = 0$ .

Set $x_i=0$ , $y_i =d(x,y)$ and $z_i = d(x,\beta_i) + d(\beta_i, y)$ for all $i \in \mathbb{N}$ . From the triangle inequality and non-negativity of the metric we now have $x_i \leq y_i \leq z_i$ for all $i \in \mathbb{N}$ .

Since $\lim_i x_i = 0$ and $\lim_i z_i = 0$ , we have by the squeeze theorem $\lim_i y_i = \lim_i d(x,y) = 0$ , therefore $d(x,y) =0$ which implies $x=y$ . We conclude that $A$ is a metric base for $(X,d)$ .

Q.E.D.

Obviously, by set inclusion we have that every $A\subseteq X$ that contains a metric base is a metric base. What about metric bases which are not dense in $X$ and also do not contain a finite metric base? Formally, we have the following question:

Question: Does there exist a metric space $(X,d)$ such that it has an infinite metric base $A$ with the following properties:

$A$ is not dense;
for all finite metric bases $B$ of $(X,d)$ we have $B \not \subseteq A$ ?