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Is single point extension of a metric space possible? Let $(X,d)$ be a metric space and $\overline{X}=X\cup \{\overline{x}\}$. Is it possible to find a metric $\overline{d}$ for which $(\overline{X},\overline{d})$ is a metric space and $\overline{d}(x)=d(x)$ for every $x\in X$?

bc78
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    I'm not sure about the answer, but why would you want such an extension? Normally if you extend a space, it's because you are trying to gain some desirable property; e.g., Alexandroff (one-point) compactification of a locally compact Hausdorff space. Is there a particular reason you want such an extension? Also, $d: X \times X \to [0,\infty)$ (and likewise for $\overline d$) so a statement like $d(x)=\overline d(x)$ doesn't have any meaning; they need two arguments. – User8128 Mar 23 '16 at 14:56
  • In probability theory, it is very common to assume that the sample space is a Borel subset of complete separable metric space, which is basically a Borel space. In probability applications extending state space by adding some absorbing states is very useful to simplify formulations and calculations. In the related textbooks, this type of extensions are quite common but rigorous explanation is missing. – bc78 Mar 23 '16 at 16:56
  • When I extend a state space $(\Omega,d)$, do I need to find a new Borel space $(\overline{\Omega},\overline{d})$, which is an extension of the previous state space, or at the beginning, should I start to define everything on the extended state space $(\overline{\Omega},\overline{d})$ and assume that the initial state space $(\Omega ,d)$ is also a subset of the extended one? – bc78 Mar 23 '16 at 16:58

2 Answers2

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It is clear that single point extensions of bounded metric spaces are possible.

Suppose that $(X, d)$ is a bounded metric space, and let $r \in \mathbb{R}_{\geq 0}$ be such that $d(x,y) \leq r$ for all $x$ and $y$ in $X$.

Let $\overline{X} = X \cup \left\{ \overline{x} \right\}$, where $\overline{x} \not\in X$. Define $$\overline{d} : \overline{X} \times \overline{X} \to \mathbb{R}$$ as follows. For $x, y \in X$, $\overline{d}(x, y) = d(x, y)$. For $x \in X$, $\overline{d}(x, \overline{x}) = \overline{d}(\overline{x}, x) = r$. Given the metric identity of indiscernables, define $\overline{d}(\overline{x},\overline{x}) = 0$.

It is easily seen that $\overline{d}$ is a metric on $\overline{X}$. It is obvious that the separation axiom holds. It is obvious that the identity of indiscernibles holds. It is obvious that the symmetry axiom holds.

It is easily verified that the triangle inequality holds. Letting $x, y, z \in X$, we have that:

1. $\overline{d}(x, z) \leq \overline{d}(x, y) + \overline{d}(y, z)$, since ${d}(x, z) \leq {d}(x, y) + {d}(y, z)$;

2. $\overline{d}(\overline{x}, z) \leq \overline{d}(\overline{x}, y) + \overline{d}(y, z)$, since $1 \leq 1 + {d}(y, z)$;

3. $\overline{d}(x, z) \leq \overline{d}(x, \overline{x}) + \overline{d}(\overline{x}, z)$, since ${d}(x, z) \leq r \leq 2r$;

John M. Campbell
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    Thank you very much for your answer. When the metric space is not bounded, can I define it on the extended real numbers? But, I guess, this is something different that I do not have a deep knowledge, unfortunately... – bc78 Mar 23 '16 at 17:01
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Well, you could,

Let $X$ be the integers and $d$ be the discrete metric $d(a, b) = 1$ if $a \ne b$ and $= 0$ if $ a = b$. The discrete metric induces the discrete topology where the open sets are all possible subsets of $X$.

Now add any non-integer to $X$ with the same (discrete) metric.

But why ?

Tom Collinge
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  • More generally, it is clear that single point extensions of bounded metric spaces are possible, as shown in my answer. – John M. Campbell Mar 23 '16 at 15:13
  • In probability theory, it is very common to assume that the sample space is a Borel subset of complete separable metric space, which is basically a Borel space. In probability applications extending state space by adding some absorbing states is very useful to simplify formulations and calculations. In the related textbooks, this type of extensions are quite common but rigorous explanation is missing. – bc78 Mar 23 '16 at 16:59
  • When I extend a state space $(\Omega,d)$, do I need to find a new Borel space $(\overline{\Omega},\overline{d})$, which is an extension of the previous state space, or at the beginning, should I start to define everything on the extended state space $(\overline{\Omega},\overline{d})$ and assume that the initial state space $(\Omega ,d)$ is also a subset of the extended one? – bc78 Mar 23 '16 at 16:59
  • Not my field (or metric space for that matter) at all. Hope someone else can help you. – Tom Collinge Mar 23 '16 at 18:01