Can someone make clear the concept of a "restricted metric"

Question

I never understood this concept of restricted metric. Consider the following theorem.

http://www.math.psu.edu/wysocki/M403/Notes403_4.pdf

I don't quite understand this concept since under usual circumstances when working with $\mathbb{R}^n$, take any subset of $\mathbb{R}^n$, then we can simply equip the subset with the Euclidean distance as in $\mathbb{R}^n$.

Here it is pointless to talk about the restriction, because it is understood that the metric is now defined on the Cartesian product of the subset. I have never seen anyone making this distinction.

Can someone demonstrate a case where $S$ a subset of $(M,d)$, where the restricted metric $d|_S$ can be quite different than the "ambient" metric $d$?

Answer 1

The key here is that the metric on $Y$ is a function $d_Y\colon Y\times Y\to [0,\infty)$. The "original" metric $d_X$ is a function $d_X\colon X\times X\to [0,\infty)$.

You have $d_Y(u,v) = d_X(u,v)$ for every $(u,v) \in Y\times Y$, but that does not mean the two functions are the same: they coincide on every point of $Y\times Y$, but they do not have the same domain. They are not formally the same mathematical object.

In particular, if $Y\subsetneq X$ and $u,v\in X\setminus Y$, then $d_X(u,v)$ is defined, but $d_Y(u,v)$ is not.

Answer 2

I don't quite understand what you mean by "distinction". Well, consider $X = \mathbb R$ with the Euclidian metric and $Y = [0,1]$. Then, for example, $[0,1]$ is open in the metric space $Y$, while it is not in $X$. The same holds for sets of the type $[0,x)$, where $x\in (0,1)$. Or: If you take $Y = (0,1)$ then $Y$ is not complete while $X$ is. So, suddenly sequences do not have limits which they had before.