A description/proof of that for metric space, difference of distances of 2 points from subset is equal to or less than distance between points
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric space.
- The reader knows a definition of distance between 2 points on metric space.
- The reader knows a definition of distance between point and subset on metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space, any subset, and any 2 points, the difference of the distances of the points from the subset is equal to or less than the distance between the points.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any metric space, \(M\), any subset, \(S \subseteq M\), and any points, \(p_1, p_2 \in M\), \(|d (p_1, S) - d (p_2, S)| \leq d (p_1, p_2)\) where \(d (\bullet, \bullet)\) denotes the distance between the arguments.
2: Proof
Suppose that \(p_1, p_2 \notin S\). Take any point, \(p'_2 \in S\). By the definition of distance between 2 points, \(d (p_1, p'_2) \leq d (p_1, p_2) + d (p_2, p'_2)\). By the definition of distance between point and subset, \(d (p_1, S) \leq d (p_1, p'_2)\) and \(d (p_2, p'_2) = d (p_2, S) + \epsilon\) where \(\epsilon \geq 0\) can be made any small positive. So, \(d (p_1, S) \leq d (p_1, p_2) + d (p_2, S) + \epsilon\). By taking the limit by \(\epsilon \rightarrow 0\), \(d (p_1, S) \leq d (p_1, p_2) + d (p_2, S)\), so, \(d (p_1, S) - d (p_2, S) \leq d (p_1, p_2)\). By the symmetricalness, also \(d (p_2, S) - d (p_1, S) \leq d (p_1, p_2)\), so \(|d (p_1, S) - d (p_2, S)| \leq d (p_1, p_2)\).
Suppose that \(p_1 \notin S\) and \(p_2 \in S\). By the definition of distance between point and subset, \(d (p_1, S) \leq d (p_1, p_2)\). As \(d (p_2, S) = 0\), \(d (p_1, S) - d (p_2, S) \leq d (p_1, p_2)\). As the left side is non-negative, \(|d (p_1, S) - d (p_2, S)| \leq d (p_1, p_2)\).
Suppose that \(p_1 \in S\) and \(p_2 \notin S\). By the symmetricalness, \(|d (p_2, S) - d (p_1, S)| \leq d (p_2, p_1)\), which implies \(|d (p_1, S) - d (p_2, S)| \leq d (p_1, p_2)\).
Suppose that \(p_1, p_2 \in S\). As \(d (p_1, S) = d (p_2, S) = 0\), \(|d (p_1, S) - d (p_2, S)| \leq d (p_1, p_2)\).
