description/proof of that for finite-dimensional vectors space with metric induced by inner product, space is complete
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of norm induced by inner product on real or complex 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 knows a definition of convergence of sequence on metric space.
- The reader admits the proposition that any countable-dimensional vectors space with any inner product has an orthonormal basis.
- The reader admits the proposition that for any net with any directed index set into any finite-dimensional real or complex vectors space with the canonical topology, the convergence exists if and only if the convergences of the components (with respect to any basis) exist, and then, the convergence is expressed with the convergences.
- The reader admits the proposition that for any finite-dimensional vectors space with any inner product, the topology induced by the inner product is the canonical topology.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space with the metric induced by any inner product, 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
\(d\): \(\in \mathbb{N}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with any inner product and the metric induced by the norm induced by the inner product
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Statements:
\(V \in \{\text{ the complete metric spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any orthonormal basis for \(V\); Step 2: take any Cauchy sequence and express it with the basis, and see that the component sequences converge; Step 3: see that the sequence converges with the component convergences.
Step 1:
There is an orthonormal basis for \(V\), \(B = \{b_1, ..., b_d\}\), by the proposition that any countable-dimensional vectors space with any inner product has an orthonormal basis.
Step 2:
Let \(s: \mathbb{N} \to V\) be any Cauchy sequence.
For each \(l \in \mathbb{N}\), \(s (l) = s (l)^j b_j\).
For each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for each \(m, n \in \mathbb{N}\) such that \(N \lt m, n\), \(dist (s (m), s (n)) \lt \epsilon\).
That means that \(\langle s (m) - s (n), s (m) - s (n) \rangle \lt \epsilon^2\).
So, \(\langle s (m)^j b_j - s (n)^j b_j, s (m)^j b_j - s (n)^j b_j \rangle = \langle (s (m)^j - s (n)^j) b_j, (s (m)^j - s (n)^j) b_j \rangle = \sum_{j \in \{1, ..., d\}} \vert s (m)^j - s (n)^j \vert^2\), because the basis is orthonormal, \(\lt \epsilon^2\).
So, for each \(l \in \{1, ..., d\}\), \(\vert s (m)^l - s (n)^l \vert^2 \le \sum_{j \in \{1, ..., d\}} \vert s (m)^j - s (n)^j \vert^2 \lt \epsilon^2\).
That means that \(s^l: \mathbb{N} \to F, n \mapsto s (n)^l\) is a Cauchy sequence.
As \(F\) as the canonical metric space is complete as is well known, \(s^l\) converges to a \(r^l\).
Step 3:
So, \(s\) converges to \(r^j b_j\) with \(V\) regarded as the canonical topological space, by the proposition that for any net with any directed index set into any finite-dimensional real or complex vectors space with the canonical topology, the convergence exists if and only if the convergences of the components (with respect to any basis) exist, and then, the convergence is expressed with the convergences.
But the canonical topological space is nothing but the topological space induced by the inner product, by the proposition that for any finite-dimensional vectors space with any inner product, the topology induced by the inner product is the canonical topology.
But the convergence on the topological space induced by (the metric induced by the norm induced by) the inner product is the convergence on the metric space, as is mentioned in Note for the definition of convergence of sequence on metric space.
So, \(s\) converges to \(r^j b_j\) with \(V\) regarded as the metric space.
So, \(V\) is complete.
