description/proof of that for metric space and subset, distance from subset map is uniformly continuous
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of distance between subset and point on metric space.
- The reader knows a definition of uniformly continuous map between metric spaces.
- The reader admits the proposition that for any metric space and any subset, the distance between the subset and any point satisfies the triangle inequality with respect to any other point.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space and any subset, the distance from the subset map is uniformly continuous.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(S\): \(\subseteq M\)
\(f\): \(: M \to \mathbb{R}, m \mapsto dist (S, m)\)
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Statements:
\(f \in \{\text{ the uniformly continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(\epsilon \in \mathbb{R}\) and see that for \(\delta = \epsilon\), \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let us take \(\delta = \epsilon\).
Let \(m \in M\) be any.
Let us see that \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\).
Let \(m' \in B_{m, \delta}\) be any.
\(dist (S, m') \le dist (S, m) + dist (m, m')\), by the proposition that for any metric space and any subset, the distance between the subset and any point satisfies the triangle inequality with respect to any other point.
That equals \(f (m') \le f (m) + dist (m, m')\).
So, \(f (m') - f (m) \le dist (m, m') \lt \delta\).
By the symmetry, \(f (m) - f (m') \lt \delta\).
So, \(\vert f (m') - f (m) \vert \lt \delta = \epsilon\).
That means that \(f (m') \in B_{f (m), \epsilon}\).
So, \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\).
As \(\delta\) is determined independent of \(m\), \(f\) is uniformly continuous.
