description/proof of that uniformly Cauchy sequence of maps from set into complete metric space converges uniformly
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of complete metric space.
- The reader knows a definition of uniformly Cauchy sequence of maps from set into metric space.
- The reader knows a definition of uniformly convergent sequence of maps from set into metric space.
- The reader knows a definition of convergence of sequence on metric space.
Target Context
- The reader will have a description and a proof of the proposition that any uniformly Cauchy sequence of maps from any set into any complete metric space converges uniformly.
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:
\(S\): \(\in \{\text{ the sets }\}\)
\(M\): \(\in \{\text{ the complete metric spaces }\}\)
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{g: S \to M\}\)
//
Statements:
\(s \in \{\text{ the uniformly Cauchy sequences }\}\)
\(\implies\)
\(s \in \{\text{ the uniformly convergent sequences }\}\)
//
2: Proof
Whole Strategy: Step 1: take the convergence, \(f: S \to M\); Step 2: see that \(s\) converges to \(f\) uniformly.
Step 1:
For each fixed \(p\), \(: J \to M, j \mapsto s (j) (p)\) is a Cauchy sequence on \(M\), because for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in J\) such that for each \(j_1, j_2 \in J\) such that \(N \lt j_1, j_2\), \(dist (s (j_1) (p), s (j_2) (p)) \lt \epsilon\).
As \(M\) is complete, it converges to \(f (p)\): the convergence is unique, by Note for the definition of convergence of sequence on metric space.
So, we have the map, \(f: S \to M\).
Step 2:
Let us see that \(s\) converges to \(f\) uniformly.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in J\) such that for each \(j_1, j_2 \in J\) such that \(N \lt j_1, j_2\), for each \(p \in S\), \(dist (s (j_1) (p), s (j_2) (p)) \lt \epsilon / 2\).
\(dist (s (j_1) (p), f (p)) \le dist (s (j_1) (p), s (j_2) (p)) + dist (s (j_2) (p), f (p)) \lt \epsilon / 2 + dist (s (j_2) (p), f (p))\).
But there is an \(N' \in J\) such that for each \(j_2 \in J\) such that \(N' \lt j_2\), \(dist (s (j_2) (p), f (p)) \lt \epsilon / 2\): \(N'\) depends on \(p\), but that does not matter, because what we need is that \(N\) does not depend on \(p\).
So, let \(j_2\) be any such that \(N, N' \lt j_2\), and \(dist (s (j_1) (p), f (p)) \lt \epsilon\).
That holds for each \(j_1 \in J\) such that \(N \lt j_1\) for each \(p \in S\).
So, \(s\) converges to \(f\) uniformly.
