description/proof of that for metric space induced by normed vectors space, if for each sequence s.t. series of norms converges, series converges, space is complete
Topics
About: vectors space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of normed vectors space.
- The reader knows a definition of metric induced by norm on real or complex vectors space.
- The reader knows a definition of complete metric space.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for the metric space induced by any normed vectors space, if for each sequence such that the series of the norms of the terms of the sequence converges, the series of the terms of the sequence converges, the space is complete.
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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), \(= \text{ the metric space induced by the following norm }\)
\(\Vert \bullet \Vert\): \(: V \to \mathbb{R}\), \(\in \{\text{ the norms }\}\)
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Statements:
\(\forall f: \mathbb{N} \to V \text{ such that } \sum_{j = 0, 1, ...} \Vert f (j) \Vert \lt \infty (\sum_{j = 0, 1, ...} f (j) \in \{\text{ the convergent series }\})\)
\(\implies\)
\(V \in \{\text{ the complete metric spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any Cauchy sequence, \(f: \mathbb{N} \to V\); Step 2: take \(N_j\) s such that for each \(m, n \in \mathbb{N}\) such that \(N_j \lt m, n\), \(\Vert f (n) - f (m) \Vert \lt 2^{- j}\) and \(N_j \lt N_{j + 1}\); Step 3: see that \(\sum_{j = 0, 1, ...} \Vert f (N_{j + 1} + 1) - f (N_j + 1) \Vert \lt \infty\); Step 4: see that \(\sum_{j = 0, 1, ...} f (N_{j + 1} + 1) - f (N_j + 1)\) converges and that the sequence, \(f (N_0 + 1), f (N_1 + 1), ...\), converges; Step 5: see that \(f\) converges.
Step 1:
Let \(f: \mathbb{N} \to V\) be any Cauchy sequence.
Step 2:
For each \(j = 0, 1, ...\), let us take any \(N_j\) such that for each \(m, n \in \mathbb{N}\) such that \(N_j \lt m, n\), \(\Vert f (n) - f (m) \Vert \lt 2^{- j}\), which is possible because \(f\) is a Cauchy sequence.
Let us take \(N_j \lt N_{j + 1}\), which is possible because if \(N_{j + 1} \le N_j\), \(N_{j + 1}\) can be just redefined to be \(N_j + 1\), because for each \(m, n \in \mathbb{N}\) such that \(N_j + 1 \lt m, n\), \(N_{j + 1} \lt m, n\), so, \(\Vert f (n) - f (m) \Vert \lt 2^{- (j + 1)}\) holds anyway.
Step 3:
Let us see that \(\sum_{j = 0, 1, ...} \Vert f (N_{j + 1} + 1) - f (N_j + 1) \Vert \lt \infty\).
\(\sum_{j = 0, 1, ...} \Vert f (N_{j + 1} + 1) - f (N_j + 1) \Vert \lt \sum_{j = 0, 1, ...} 2^{- j}\), because \(N_j \lt N_{j + 1} + 1, N_j + 1\).
\(= 1 / (1 - 1 / 2) \lt \infty\), because \(\sum_{j = 0, 1, ..., n} 2^{- j} = 1 + 1 / 2 + (1 / 2)^2 + ... + (1 / 2)^n = (1 - 1 / 2^{n + 1}) / (1 - 1 / 2)\), which limits to \(1 / (1 - 1 / 2)\) by \(n \to \infty\).
Step 4:
By the supposition of this proposition, \(\sum_{j = 0, 1, ...} f (N_{j + 1} + 1) - f (N_j + 1)\) converges.
But \(\sum_{j = 0, 1, ..., n} f (N_{j + 1} + 1) - f (N_j + 1) = (f (N_1 + 1) - f (N_0 + 1)) + (f (N_2 + 1) - f (N_1 + 1)) + ... + (f (N_n + 1) - f (N_{n- 1} + 1)) + (f (N_{n + 1} + 1) - f (N_n + 1)) = f (N_{n + 1} + 1) - f (N_0 + 1)\), so, the sequence, \(f (N_1 + 1) - f (N_0 + 1), f (N_2 + 1) - f (N_0 + 1), ...\), converges to a \(v\), so, the sequence, \(f (N_1 + 1), f (N_2 + 1), ...\), converges to \(f (N_0 + 1) + v\): \(\Vert f (N_n + 1) - f (N_0 + 1) - v \Vert = \Vert f (N_n + 1) - (f (N_0 + 1) + v) \Vert\), so, the estimates for \(f (N_1 + 1) - f (N_0 + 1), f (N_2 + 1) - f (N_0 + 1), ...\) to converge to \(v\) are the same with the estimates for \(f (N_1 + 1), f (N_2 + 1), ...\) to converge to \(f (N_0 + 1) + v\).
Step 5:
As \(f (N_1 + 1), f (N_2 + 1), ...\) is a subsequence of \(f\), \(f\) converges, by 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.
So, as each Cauchy sequence converges, \(V\) is complete.
