Does every metric on a non empty set have an extension to an arbitrary super set? Let $\phi \ne X \subseteq Y$ , let $d$ be a metric on $X$ , then does there exist a metric $d'$ on $Y$ such that $d(x,y)=d'(x,y) , \forall x, y \in X$ ? What if we also assume that the metric $d$ on $X$ is bounded ? Please help . Thanks in advance.
 A: If $Y$ is just an arbitrary set containing $X$, then you can equip $Y \setminus X$ with any metric you like (say, a discrete one) and join the spaces via some distinguished element.
Let $d''$ be a metric on $Y \setminus X$ and let $x_0$ and $y_0$ be any elements of $X$ and $Y \setminus X$ respectively. Then
$$
d'(\alpha,\beta) = 
  \begin{cases}
    d(\alpha,\beta) & \text{ if }\alpha, \beta \in X \\
    d''(\alpha,\beta) & \text{ if } \alpha, \beta \in Y \setminus X \\
    d(\alpha, x_0) + 1 + d(y_0, \beta) & \text{ if } \alpha \in X, \beta \in Y\setminus X \\
    d(\alpha, y_0) + 1 + d(x_0, \beta) & \text{ if } \alpha \in Y\setminus X, \beta \in X
  \end{cases}
$$
is a metric on $Y$.
I hope this helps $\ddot\smile$
A: Certainly the answer is yes if $d$ is bounded. Assume that $d(x_1,x_2)\le M$ for all $x_1,x_2\in X$. For all $y_1,y_2\in Y$, define
$$
d'(y_1,y_2) = \begin{cases}
d(y_1,y_2), &\text{if } y_1\in X \text{ and } y_2\in X, \\
0, &\text{if } y_1=y_2\notin X, \\
M, &\text{if } y_1\notin X \text{ or } y_2\notin X.
\end{cases}
$$
Then one can check that $d'$ is a metric.
