description/proof of that for metric space, Cauchy sequence into metric space, and positive number, there is subsequence s.t. distance between \(j\)-th element and '\(j + 1\)'-th element is smaller than number to '\(j + 1\)'-th power
Topics
About: metric space
The table of contents of this article
Starting Context
- 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 for any metric space, any Cauchy sequence into the metric space, and any positive number, there is a subsequence such that for each \(j\), the distance between the \(j\)-th element and the '\(j + 1\)'-th element is smaller than the number to the '\(j + 1\)'-th power.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(s\): \(: \mathbb{N} \to M\), \(\in \{\text{ the Cauchy sequences }\}\)
\(p\): \(\in \mathbb{R}\) such that \(0 \lt p\)
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Statements:
\(\exists f: \mathbb{N} \to \mathbb{N} \text{ such that } \forall j, l \in \mathbb{N} \text{ such that } j \lt l (f (j) \lt f (l)) (\forall j \in \mathbb{N} (dist (s \circ f (j + 1), s \circ f (j)) \lt p^{j + 1}))\)
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2: Note
Usually, \(p\) is taken such that \(0 \lt p \lt 1\), but that is not required.
A motivation for this proposition is that while the proposition that for any Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point holds, if \(s \circ f\) converges, \(s\) will converge, and \(s \circ f\) may be easier to be proved to converge.
3: Proof
Whole Strategy: define \(f\) inductively; Step 1: define \(f (0)\) and \(f (1)\); Step 2: suppose that \(f (0), ..., f (n - 1)\) have been defined, and define \(f (n)\); Step 3: see that \(f\) satisfies the conditions for this proposition.
Step 1:
Let us define \(f (0)\) and \(f (1)\).
There is an \(N_0 \in \mathbb{N}\) such that for each \(j, l \in \mathbb{N}\) such that \(N_0 \lt j, l\), \(dist (s (l), s (j)) \lt p^{0 + 1}\), because \(s\) is Cauchy.
Let \(f (0) := N_0 + 1\).
There is an \(N_1 \in \mathbb{N}\) such that for each \(j, l \in \mathbb{N}\) such that \(N_1 \lt j, l\), \(dist (s (l), s (j)) \lt p^{1 + 1}\), because \(s\) is Cauchy.
Let \(f (1) := max (f (0), N_1) + 1\).
\(f (0) \lt f (1)\).
Step 2:
Let us suppose that \(f (0), ..., f (n - 1)\) have been defined with \(f (0) \lt ... \lt f (n - 1)\).
There is an \(N_n \in \mathbb{N}\) such that for each \(j, l \in \mathbb{N}\) such that \(N_n \lt j, l\), \(dist (s (l), s (j)) \lt p^{n + 1}\), because \(s\) is Cauchy.
Let \(f (n) := max (f (n - 1), N_n) + 1\).
\(f (n - 1) \lt f (n)\), so, \(f (0) \lt ... \lt f (n)\).
Step 3:
Let \(j, l \in \mathbb{N}\) be any such that \(j \lt l\).
\(f (0) \lt ... \lt f (l)\), so, \(f (j) \lt f (l)\).
Let \(j \in \mathbb{N}\) be any.
As \(N_j \lt f (j) \lt f (j + 1)\), \(dist (s \circ f (j + 1), s \circ f (j)) \lt p^{j + 1}\).
So, \(f\) satisfies the conditions.
