Metric space extension 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 $x\in X$?
 A: 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 ${d}(x, z) \leq r \leq 2r$; 
A: 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 ?
