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.
Asked
Active
Viewed 148 times
2
-
Are you assuming that $Y$ is a topological space, and $X$ is a subspace that happens to be metrizable? Or is $Y$ simply a set containing $X$? – Brian M. Scott May 01 '16 at 15:59
-
@BrianM.Scott : $Y$ is simply any set containing arbitrary $X$ and $d$ be any metric on $X$ – May 01 '16 at 16:05
2 Answers
6
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 $\ddot\smile$
dtldarek
- 37,381
2
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.
Greg Martin
- 78,820