A description/proof of that 2 metrics with condition with each other define same topology
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of topology.
- The reader knows a definition of metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any set, any 2 metrics on the set with a condition with each other define the same topology.
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 set, \(S\), any metrics, \(dist_1\) and \(dist_2\), on \(S\) that satisfy the condition that for any point, \(p \in S\), and any number, \(0 \lt \epsilon\), there is a number, \(0 \lt \delta\), such that for any \(p'\) that satisfies \(dist_1 (p, p') \lt \delta\), \(dist_2 (p, p') \lt \epsilon\) and for any \(p'\) that satisfies \(dist_2 (p, p') \lt \delta\), \(dist_1 (p, p') \lt \epsilon\) define the same topology.
2: Proof
Suppose that \(U\) is open with respect to \(dist_1\). At any point, \(p \in S\), there is an open ball with respect to \(dist_1\), \(B_{1-p-\epsilon} \subseteq U\). Let us show that there is a \(0 \lt \delta\) such that \(B_{2-p-\delta} \subseteq B_{1-p-\epsilon}\) where \(B_{2-p-\delta}\) is the open ball with respect to \(dist_2\). Let us take \(\delta\) as satisfies the condition with respect to \(p\) and \(\epsilon\). For any \(p' \in B_{2-p-\delta}\), \(dist_2 (p, p') \lt \delta\), so, \(dist_1 (p, p') \lt \epsilon\), so, \(p' \in B_{1-p-\epsilon}\), so, \(B_{2-p-\delta} \subseteq B_{1-p-\epsilon} \subseteq U\). So, \(U\) is open with respect to \(dist_2\).
By the symmetry, any \(U\) that is open with respect to \(dist_2\) is open with respect to \(dist_1\).
