description/proof of that continuous map from complete metric space into metric space maps Cauchy sequence to Cauchy sequence
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 continuous map.
- The reader knows a definition of Cauchy sequence on metric space.
- The reader admits the proposition that for any sequence of points on any metric space, the sequence is Cauchy if and only if for each \(\epsilon\), there is an \(N\) such that the distance between the \((N + 1)\)-th point and each subsequent point is smaller than \(\epsilon\).
Target Context
- The reader will have a description and a proof of the proposition that any continuous map from any complete metric space into any metric space maps any Cauchy sequence to a Cauchy sequence.
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:
\(T_1\): \(\in \{\text{ the complete metric spaces }\}\)
\(T_2\): \(\in \{\text{ the metric spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(s\): \(: \mathbb{N} \setminus \{0\} \to T_1\), \(\in \{\text{ the Cauchy sequences }\}\)
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Statements:
\(f \circ s: \mathbb{N} \setminus \{0\} \to T_2 \in \{\text{ the Cauchy sequences }\}\)
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2: Proof
Whole Strategy: Step 1: take a convergence of \(s\), \(p \in T_1\), and \(f (p) \in T_2\); Step 2: see that \(f \circ s\) is a Cauchy sequence.
Step 1:
Let us take a convergence of \(s\), \(p \in T_1\), which is possible because \(s\) is a Cauchy sequence and \(T_1\) is complete.
Then, \(f (p) \in T_2\) exists.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let us take the \(\epsilon / 2\)-'open ball' around \(f (p)\), \(B_{f (p), \epsilon / 2} \subseteq T_2\).
As \(f\) is continuous, there is a \(\delta\)-'open ball' around \(p\), \(B_{p, \delta}\), such that \(f (B_{p, \delta}) \subseteq B_{f (p), \epsilon / 2}\).
As \(s\) converges to \(p\), there is an \(N \in \mathbb{N} \setminus \{0\}\) such that for each \(j \in \mathbb{N} \setminus \{0\}\) such that \(N \lt j\), \(s (j) \in B_{p, \delta}\).
So, for each \(j \in \mathbb{N} \setminus \{0\}\) such that \(N \lt j\), \(f \circ s (j) \in f (B_{p, \delta}) \subseteq B_{f (p), \epsilon / 2}\).
Especially, \(f \circ s (N + 1) \in B_{f (p), \epsilon / 2}\).
\(dist (f \circ s (j), f \circ s (N + 1)) \le dist (f \circ s (j), f (p)) + dist (f (p), f \circ s (N + 1)) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
So, by the proposition that for any sequence of points on any metric space, the sequence is Cauchy if and only if for each \(\epsilon\), there is an \(N\) such that the distance between the \((N + 1)\)-th point and each subsequent point is smaller than \(\epsilon\), \(f \circ s\) is a Cauchy sequence.
3: Note
The logic requires that \(T_1\) is complete.
If we take an arbitrary \(N\) and \(B_{f \circ s (N + 1), \epsilon}\), there will be a \(B_{s (N + 1), \delta}\) such that \(f (B_{s (N + 1), \delta}) \subseteq B_{f \circ s (N + 1), \epsilon}\), but there will be no guarantee that for each \(N \lt j\), \(s (j) \in B_{s (N + 1), \delta}\), because \(N\) has not been chosen for that purpose, and \(f \circ s (j) \in B_{f \circ s (N + 1), \epsilon}\) will not necessarily follow; certainly, we will be able to choose an \(N'\) such that for each \(N' \lt j\), \(s (j) \in B_{s (N' + 1), \delta}\), but \(f (B_{s (N' + 1), \delta}) \subseteq B_{f \circ s (N + 1), \epsilon}\) will not be guaranteed.
