description/proof of that metric is continuous w.r.t. topology induced by metric
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of product topology.
- The reader knows a definition of continuous map.
Target Context
- The reader will have a description and a proof of the proposition that any metric is continuous with respect to the topology induced by the metric.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the metric spaces }\}\), with metric, \(dist: T \times T \to \mathbb{R}\) with the topology induced by \(dist\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
//
Statements:
\(dist \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(p = (p^1, p^2) \in T \times T\), any neighborhood of \(dist (p^1, p^2)\), \(N_{dist (p^1, p^2)} \subseteq \mathbb{R}\), and an open ball around \(dist (p^1, p^2)\), \(B_{dist (p^1, p^2), \epsilon} \subseteq \mathbb{R}\), such that \(B_{dist (p^1, p^2), \epsilon} \subseteq N_{dist (p^1, p^2)}\); Step 2: take the open neighborhood of \(p\), \(B_{p^1, \epsilon / 2} \times B_{p^2, \epsilon / 2} \subseteq T \times T\), and see that \(dist (B_{p^1, \epsilon / 2} \times B_{p^2, \epsilon / 2}) \subseteq B_{dist (p^1, p^2), \epsilon} \subseteq N_{dist (p^1, p^2)}\).
Step 1:
Let \(p = (p^1, p^2) \in T \times T\) be any.
Let \(N_{dist (p^1, p^2)} \subseteq \mathbb{R}\) be any neighborhood of \(dist (p^1, p^2)\).
There is an open ball around \(dist (p^1, p^2)\), \(B_{dist (p^1, p^2), \epsilon} \subseteq \mathbb{R}\), such that \(B_{dist (p^1, p^2), \epsilon} \subseteq N_{dist (p^1, p^2)}\).
Step 2:
Let us take the open neighborhood of \(p\), \(B_{p^1, \epsilon / 2} \times B_{p^2, \epsilon / 2} \subseteq T \times T\), which is indeed open on \(T \times T\) by the definition of topology induced by metric and the definition of product topology.
For any point, \(p' = (p'^1, p'^2) \in B_{p^1, \epsilon / 2} \times B_{p^2, \epsilon / 2}\), \(dist (p'^1, p'^2) \le dist (p'^1, p^1) + dist (p^1, p'^2) \le dist (p'^1, p^1) + dist (p^1, p^2) + dist (p^2, p'^2) \lt \epsilon / 2 + dist (p^1, p^2) + \epsilon / 2\), so, \(dist (p'^1, p'^2) - dist (p^1, p^2) \lt \epsilon\); \(dist (p^1, p^2) \le dist (p^1, p'^1) + dist (p'^1, p^2) \le dist (p^1, p'^1) + dist (p'^1, p'^2) + dist (p'^2, p^2) \lt \epsilon / 2 + dist (p'^1, p'^2) + \epsilon / 2\), so, \(dist (p^1, p^2) - dist (p'^1, p'^2) \lt \epsilon\); so, \(\vert dist (p'^1, p'^2) - dist (p^1, p^2) \vert \lt \epsilon\).
That means that \(dist (B_{p^1, \epsilon / 2} \times B_{p^2, \epsilon / 2}) \subseteq B_{dist (p^1, p^2), \epsilon} \subseteq N_{dist (p^1, p^2)}\). So, \(dist\) is continuous.
