description/proof of that for Cauchy sequence on metric space, if subsequence converges to point, sequence converges to the point
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 Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point.
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:
\(T\): \(\in \{\text{ the metric spaces }\}\)
\(f\): \(: \mathbb{N} \to T\), \(\in \{\text{ the Cauchy sequences }\}\)
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Statements:
\(\exists f': \mathbb{N} \to \mathbb{N} \text{ such that } \forall j, k \in \mathbb{N} \text{ such that } j \lt k (f' (j) \lt f' (k)) (f \circ f' \to t \in T, \text{ which means that } f \circ f' \text{ converges to } t)\)
\(\implies\)
\(f \to t \in T\), which means that \(f\) converges to \(t\)
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2: Proof
Whole Strategy: Step 1: take the open ball around \(t\), \(B_{t, \epsilon / 2}\), and take an \(N \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N \lt j\), \(f \circ f' (j) \in B_{t, \epsilon / 2}\); Step 2: take an \(N' \in \mathbb{N}\) such that for each \(j, k \in \mathbb{N}\) such that \(N' \lt j, k\), \(dist (f (j), f (k)) \lt \epsilon / 2\); Step 3: take \(N'' := max (N, N')\), and see that for each \(j \in \mathbb{N}\) such that \(N'' \lt j\), \(dist (t, f (j)) \lt \epsilon\).
Step 1:
Let us take the open ball around \(t\), \(B_{t, \epsilon / 2}\).
Let us take an \(N \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N \lt j\), \(f \circ f' (j) \in B_{t, \epsilon / 2}\), which is possible because \(f \circ f'\) converges to \(t\).
Step 2:
Let us take an \(N' \in \mathbb{N}\) such that for each \(j, k \in \mathbb{N}\) such that \(N' \lt j, k\), \(dist (f (j), f (k)) \lt \epsilon / 2\), which is possible because \(f\) is a Cauchy sequence.
Step 3:
Let us take \(N'' := max (N, N')\).
For each \(j \in \mathbb{N}\) such that \(N'' \lt j\), \(dist (t, f (j)) \leq dist (t, f \circ f' (j)) + dist (f \circ f' (j), f (j)) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\), because while \(f \circ f' (j) = f (f' (j))\), \(j \leq f' (j)\) and \(N' \le N'' \lt f' (j), j\).
So, \(f\) converges to \(t\).
