description/proof of that finite-dimensional real vectors space topology defined based on coordinates space does not depend on choice of basis
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of canonical topology for finite-dimensional real vectors space.
- The reader admits the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
- The reader admits the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional real vectors space, the topology defined by the Euclidean topology of the coordinates space based on any basis does not depend on the choice of basis.
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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional real vectors spaces }\}\)
\(B\): \(= \{b_1, ..., b_d\} \subseteq V\), \(\in \{\text{ the bases for } V\}\)
\(B'\): \(= \{b'_1, ..., b'_d\} \subseteq V\), \(\in \{\text{ the bases for } V\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean topological space }\)
\(f\): \(: V \to \mathbb{R}^d\), \(v = v^j b_j \mapsto (v^1, ..., v^d)\)
\(f'\): \(: V \to \mathbb{R}^d\), \(v = v'^j b'_j \mapsto (v'^1, ..., v'^d)\)
\(O\): \(= \{U \subseteq V \vert f (U) \in \text{ the topology of } \mathbb{R}^d\}\)
\(O'\): \(= \{U \subseteq V \vert f' (U) \in \text{ the topology of } \mathbb{R}^d\}\)
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Statements:
\(O = O'\)
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2: Proof
Whole Strategy: Step 1: see that \(f' = \phi \circ f\), where \(\phi: \mathbb{R}^d \to \mathbb{R}^d\) is the coordinates transition map, and see that \(\phi\) is a homeomorphism; Step 2: conclude the proposition.
Step 1:
\(f\) is a bijection.
\(f'\) is a bijection.
\(f' = \phi \circ f\), where \(\phi: \mathbb{R}^d \to \mathbb{R}^d\) is the bijective coordinates transition map, where there is a constant invertible matrix, \(M\), such that \(\phi: (r^1, ..., r^d)^t \mapsto (M^1_j r^j, ..., M^d_j r^j)\), by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
Let \(\mathbb{R}^d\) have the Euclidean atlas being the Euclidean \(C^\infty\) manifold.
The formula for \(\phi\) means that \(\phi\) is continuous in the norm sense for the coordinates function.
So, by the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, \(\phi\) is continuous in the topological sense.
\(\phi^{-1}: \mathbb{R}^d \to \mathbb{R}^d, (r^1, ..., r^d) \mapsto ({M^{-1}}^1_j r^j, ..., {M^{-1}}^d_j r^j)\).
So, \(\phi^{-1}\) is continuous, likewise.
So, \(\phi\) is a homeomorphism.
Step 2:
So, for each \(U \in O\), \(f' (U) = \phi \circ f (U) \subseteq \mathbb{R}^d\) is open, because \(f (U) \subseteq \mathbb{R}^d\) is open and \(\phi\) is a homeomorphism.
That means \(U \in O'\).
By symmetry, for each \(U' \in O'\), \(U' \in O\).
So, \(O = O'\).
