description/proof of that for sequence of points on metric space, sequence is Cauchy iff for each \(\epsilon\), there is \(N\) s.t. distance between \((N + 1)\)-th point and each subsequent point is smaller than \(\epsilon\)
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 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 \(\epsilon\).
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:
\(T\): \(\in \{\text{ the metric spaces }\}\)
\(s\): \(: \mathbb{N} \setminus \{0\} \to T\)
//
Statements:
\(s \in \{\text{ the Cauchy sequences }\}\)
\(\iff\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists N \in \mathbb{N} \setminus \{0\} (\forall j \in \mathbb{N} \setminus \{0\} \text{ such that } N \lt j (dist (s (N + 1), s (j)) \lt \epsilon)))\)
//
2: Proof
Whole Strategy: Step 1: suppose that there is such an \(N\); Step 2: see that \(s\) satisfies the Cauchy condition; Step 3: suppose that \(s\) satisfies the Cauchy condition; Step 4: see that there is such an \(N\).
Step 1:
Let us suppose that there is such an \(N\).
Step 2:
Let \(\epsilon\) be any.
For \(\epsilon / 2\), there is an \(N\) such that for each \(j\) such that \(N \lt j\), \(dist (s (N + 1), s (j)) \lt \epsilon / 2\).
For each \(l, m\) such that \(N \lt l, m\), \(dist (s (l), s (m)) \le dist (s (l), s (N + 1)) + dist (s (N + 1), s (m)) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\), which means that \(s\) satisfies the Cauchy condition.
Step 3:
Let us suppose that \(s\) satisfies the Cauchy condition.
Let \(\epsilon\) be any.
There is an \(N\) such that for each \(l, m\) such that \(N \lt l, m\), \(dist (s (l), s (m)) \lt \epsilon\).
Taking \(l = N + 1\), \(dist (s (N + 1), s (m)) \lt \epsilon\), for each \(m\) such that \(N \lt m\).
