description/proof of that Lipschitz map between metric spaces 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 Lipschitz map between metric spaces.
- The reader knows a definition of Cauchy sequence on metric space.
Target Context
- The reader will have a description and a proof of the proposition that any Lipschitz map between any metric spaces 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:
\(T_1\): \(\in \{\text{ the metric spaces }\}\)
\(T_2\): \(\in \{\text{ the metric spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the Lipschitz maps }\}\), with \(L\)
\(s\): \(: \mathbb{N} \setminus \{0\} \to T_1\), \(\in \{\text{ the Cauchy sequences }\}\)
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Statements:
\(f \circ s \in \{\text{ the Cauchy sequences }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(\epsilon\) and any \(N\) such that for each \(N \lt j, l\), \(dist (s (j), s (l)) \lt \epsilon / L\); Step 2: see that \(dist (f \circ s (j), f \circ s (l)) \lt \epsilon\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
As \(s\) is Cauchy, there is an \(N \in \mathbb{N} \setminus \{0\}\) such that for each \(j, l \in \mathbb{N} \setminus \{0\}\) such that \(N \lt j, l\), \(dist (s (j), s (l)) \lt \epsilon / L\).
Step 2:
As \(f\) is Lipschitz, \(dist (f \circ s (j), f \circ s (l)) \le L dist (s (j), s (l)) \lt L \epsilon / L = \epsilon\).
So, \(f \circ s\) is Cauchy.
