Virtual point of a metric space.

Question

For a metric space $X$, consider a fixed point $z$. Let $\delta_z(x) = d(z,x)$ denote the distance from a point $x$ to $z$. Then $\delta_z$ (or, more simply $\delta$) is a function from $X$ to $\mathbb R$ measuring the distance of every point in $X$ from $z$.

This function satisfies

1. $\delta(x)-\delta(y)\leq d(x,y)\leq \delta(x)+\delta(y)$,
2. $\inf_{x\in X}\delta(x)=0$, and
3. $\delta(z)=0.$

Now if we consider a function $u$ satisfying these properties except the third one i.e.

1. $u(a)-u(b)\leq d(a,b)\leq u(a)+u(b)$,
2. $\inf u(X)=0$, but
3. $0\notin u(X)$.

Then we can think of a virtual point $z$ which is not actually in $X$. If it existed, then by setting $u(z) = 0$, the function $u$ would measure distance from $z$ to each point in $X$. Actually it is related to completeness.

We say a metric space has a virtual point if such a function $u:X\to \mathbb R$ exists. To make it clear let us assume that such a function exists. Then we can construct a sequence $(x_n)$ in $X$ such that $0<u(x_n)<1/n$ for each $n$, i.e. $u(x_n)\to 0$ but $u(x_n)$ is never equal to $0$. Then this sequence $(x_n)$ shall be Cauchy from the first property of $u$. But it will not be convergent because then $u(x)=0$ for $x=\lim x_n$. Moreover, we can say, a metric space is complete if and only it has no virtual points.Is there any related link available online on the topic virtual points and completeness because I have yet not found one.Secondly,are there any other interesting aspect that I have missed here.