description/proof of that for bijective complex-conjugate-linear 'normed vectors space' isometry, if domain is complete, codomain is complete
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of bijection.
- The reader knows a definition of complex-conjugate-linear map.
- The reader knows a definition of 'normed vectors space' isometry.
- 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 complex-conjugate-linear bijection, the inverse is complex-conjugate-linear.
- The reader admits the proposition that for any bijective 'normed vectors space' isometry, the inverse is a 'normed vectors space' isometry.
- The reader admits the proposition that any complex-conjugate-linear 'normed vectors space' isometry maps any Cauchy sequence to a Cauchy sequence.
- The reader admits the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.
- The reader admits the proposition that the topological space induced by any metric is Hausdorff.
Target Context
- The reader will have a description and a proof of the proposition that for any bijective complex-conjugate-linear 'normed vectors space' isometry, if the domain is complete, the codomain 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
\(V_1\): \(\in \{\text{ the } F \text{ normed vectors spaces }\}\), with the metric induced by the norm
\(V_2\): \(\in \{\text{ the } F \text{ normed vectors spaces }\}\), with the metric induced by the norm
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the bijective complex-conjugate-linear 'normed vectors space' isometries }\}\)
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Statements:
\(V_1 \in \{\text{ the complete metric spaces }\}\)
\(\implies\)
\(V_2 \in \{\text{ the complete metric spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any Cauchy sequence on \(V_2\), \(s\), and see that \(f^{-1} \circ s\) is a Cauchy sequence on \(V_1\) and take the convergence of \(f^{-1} \circ s\), \(v\); Step 2: see that \(f (v)\) is the convergence of \(s\).
Step 1:
Let \(s: \mathbb{N} \to V_2\) be any Cauchy sequence.
\(f^{-1}\) is complex-conjugate-linear, by the proposition that for any complex-conjugate-linear bijection, the inverse is complex-conjugate-linear.
\(f^{-1}\) is 'normed vectors space' isometric, by the proposition that for any bijective 'normed vectors space' isometry, the inverse is a 'normed vectors space' isometry.
\(f^{-1} \circ s\) is a Cauchy sequence on \(V_1\), by the proposition that any complex-conjugate-linear 'normed vectors space' isometry maps any Cauchy sequence to a Cauchy sequence.
As \(V_1\) is complete, \(f^{-1} \circ s\) converges to a \(v \in V_1\), which is in fact the unique convergence, by the proposition that the topological space induced by any metric is Hausdorff and the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.
Step 2:
Let us see that \(f (v)\) is the convergence of \(s\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(dist (f^{-1} \circ s (n), v) = \Vert f^{-1} \circ s (n) - v \Vert \lt \epsilon\).
But \(\Vert s (n) - f (v) \Vert = \Vert f (f^{-1} \circ s (n)) - f (v) \Vert = \Vert f (f^{-1} \circ s (n) - v) \Vert\), because \(f\) is complex-conjugate-linear, \(= \Vert f^{-1} \circ s (n) - v \Vert\), because \(f\) is isometric, so, \(\Vert s (n) - f (v) \Vert \lt \epsilon\).
That means that \(s\) converges to \(f (v)\).
So, \(V_2\) is complete.
