Suppose $(X, d)$ is a metric space and $u \colon X → \mathbb{R}$. Then $u$ will be called a virtual point of $X$ if, and only if, $u$ satisfies the following three conditions:

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

What is the intuition behind such a definition? This is from Mícheál Ó Searcóid’s book where they have proposed an equivalent condition of $X$ having no virtual points and $X$ being complete. Can someone please explain.

  • 2
    $\begingroup$ The intuition behind that definition? It is simply a notational convenience for $d(z,x)$, no more, no less. Judging by your question title, you might be asking what is the link between $\delta$ and virtual points; if so, copy the relevant bits of the book into your question and clarify $\endgroup$
    – FShrike
    Feb 15, 2022 at 9:06
  • $\begingroup$ @FShrike can you check now? $\endgroup$
    – Antimony
    Feb 15, 2022 at 9:16
  • 3
    $\begingroup$ The third axiom: $u(z)\neq0$. As $z$ is undefined, can I take that to mean that $u$ is everywhere nonzero? $\endgroup$
    – FShrike
    Feb 15, 2022 at 9:24
  • $\begingroup$ @FShrike: Yes, the conditions are clearly meant to apply for all $x, y, z \in X$. No idea why the author chose to explicitly quantify the infimum in condition 2 but not the other conditions. $\endgroup$ Feb 15, 2022 at 18:30

2 Answers 2


This function has the properties of a function of the form $u: x\mapsto d(a,x)$ for some mysterious point $a$ which does not actually exist (thanks to condition 3; if you remove condition 3 then you can actually take $u(x)=d(a,x)$ for some $a\in X$).

What does it mean that $a$ does not exist? Well, it might be that $X\subset Y$, and $a\in Y$ but $a\not\in X$, so as far as $X$ is concerned $a$ is not actually a point.

For instance, you can define the function $u(x)=|x-\sqrt{2}|$ on $\mathbb{Q}$, which will be a virtual point of $\mathbb{Q}$, because $\mathbb{Q}$ does not see $\sqrt{2}\in \mathbb{R}$. And you can do that for any $a\in \mathbb{R}$ instead of $\sqrt{2}$.

So the intuition is that this type of function will be able to "imitate" the presence of points in a bigger space than $X$, while not actually knowing what this space looks like. So they are sort of "potential" points, or... "virtual" points. (Of course once you know about the completion, you can actually construct the bigger space.)

EDIT: I initially just wanted to provide general intuition, but since the answer has gained attention I'll provide more details. What does each axiom say?

Axiom 1 is the one that makes the function $u$ behave like $x\mapsto d(a,x)$ for some $a$ in a bigger space $Y$. Precisely, given a metric space $X$ and a function $u:X\to \mathbb{R}$ satisfying Axiom 1, one can simply define a bigger space $Y=X\cup \{a\}$ (with some symbol $a\not\in X$) and extend the metric of $X$ to $Y$ by $d(x,a)=d(a,x):= u(x)$ for each $x\in X$ (and obviously $d(a,a)=0$). Axiom 1 exactly says that this extended $d$ satisfies the triangular inequality.

This being said, with just Axiom 1, this extended $d$ could be just a quasi-metric, and not an actual metric. If there were some $x\in X$ such that $u(x)=0$, then with this construction, we would get $d(x,a)=0$, but $x\neq a$, so the separation axiom would not be satisfied! This is because when Axiom 3 is not satisfied, then $u$ already corresponds to an actual point of $X$, so we are sort of adding a second copy of a point which already exists, and those two copies get indistinguishable by the metric. So the extended $d$ to is an actual metric exactly when Axiom 3 is satisfied.

(Another way of doing things would be to say that when Axiom 3 is not satisfied, then we don't add a new point to $X$, we just take $Y=X$, and $a\in Y$ is the unique point such that $u(a)=0$. It has to be unique, since if $u(a)=u(b)=0$, then $d(a,b)\leqslant u(a)+u(b)=0$ so $a=b$.)

Finally, Axiom 2 says that this $a$ is not some isolated point in this bigger $Y$. More precisely, it is a limit point of $X$ in $Y$: $a\in \overline{X}\subset Y$. This is useful if we want to say that those virtual points are actually points in the completion, since that is exactly what the completion of a metric space is: we just add all the missing limit points of $X$.

  • 5
    $\begingroup$ This is such a wonderful answer.Thank you.. $\endgroup$
    – Antimony
    Feb 15, 2022 at 9:39
  • $\begingroup$ Maybe add that the extra point is forced to be "approximable" by the original points due to Item 2. So it can't just be any extra point, it has to come from the completion. $\endgroup$
    – Arno
    Feb 15, 2022 at 18:10
  • 2
    $\begingroup$ This is indeed a good answer, but I notice that you haven't said anything about condition 2, which seems to essentially require that the "virtual point" isn't isolated from the actual points. Which, of course, is kind of an important thing to require if your goal is to draw a connection between the absence of virtual points and completeness. $\endgroup$ Feb 15, 2022 at 18:10
  • $\begingroup$ In your second paragraph it might be helpful to add one clause to your list of "might be"s, namely that $a \in \overline{X}$. $\endgroup$
    – Lee Mosher
    Feb 15, 2022 at 22:17

Conditions 1 and 3 in the definition you quote are equivalent to saying that, if you added an extra point $a \notin X$ to $X$ and defined $$d(a,x) = d(x,a) = u(x)$$ for all $x \in X$, and $d(a,a) = 0$, then the set $X \cup \{a\}$, equipped with the expanded distance function $d$ thus defined, would still be a metric space. Specifically, condition 1 ensures that the expanded metric $d$ still satisfies the triangle inequality, while condition 3 ensures that $d(a,x) ≠ 0$ for any pre-existing point $x \in X$.

As for condition 2, what it amounts to is requiring that the extra point $a$ thus added will not be an isolated point, but can be approached arbitrarily closely by existing points in $X$. This additional condition is essential if your goal is to define the absence of "virtual points" as equivalent to completeness, since without it all metric spaces would have infinitely many virtual points.

As an example of how to construct an "isolated virtual point" violating condition 2, take any point $z \in X$ and any real number $c > 0$ and let $u(x) = d(x, z) + c$ for all $x \in X$. Clearly any such function $u$ will satisfy conditions 1 and 3, but cannot satisfy condition 2.

(The "geometric intuition" behind this definition is that the new point $a$ is placed at distance $c$ from $z$, and its distance from other points in $X$ is defined by assuming that you can only reach $a$ from $X$ by going through $z$. Obviously other arrangements are possible too: for example, we could make the new point $a$ reachable via $n$ points in $X$ instead of just one, i.e. pick $z_1, \dots, z_n \in X$ and $c_1, \dots, c_n \in \mathbb R^+$, making sure that $c_i + c_j ≥ d(z_i, z_j)$ for all $1 ≤ i, j ≤ n$ so that condition 1 will still be satisfied, and let $u(x) = \min_{1 ≤ i ≤ n}(d(x, z_i) + c_i)$.)


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