125: Finite-Dimensional Real Vectors Space Topology Defined Based on Coordinates Space Does Not Depend on Choice of Basis

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description/proof of that finite-dimensional real vectors space topology defined based on coordinates space does not depend on choice of basis

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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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 $C^{\mathrm{\infty }}$ manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$V$: $\in \{\text{ the }d\text{ -dimensional real vectors spaces }\}$
$B$: $=\{b_1,...,b_d\}\subseteq V$, $\in \{\text{ the bases for }V\}$
$B^{\prime }$: $=\{b_1^{\prime },...,b_d^{\prime }\}\subseteq V$, $\in \{\text{ the bases for }V\}$
${\mathbb{R}}^d$: $=\text{ the Euclidean topological space }$
$f$: $:V\to {\mathbb{R}}^d$, $v=v^j{b}_j\mapsto (v^1,...,v^d)$
$f^{\prime }$: $:V\to {\mathbb{R}}^d$, $v=v^{\prime j}{b}_j^{\prime }\mapsto (v^{\prime 1},...,v^{\prime d})$
$O$: $=\{U\subseteq V|f(U)\in \text{ the topology of }{\mathbb{R}}^d\}$
$O^{\prime }$: $=\{U\subseteq V|f^{\prime }(U)\in \text{ the topology of }{\mathbb{R}}^d\}$
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Statements:
$O=O^{\prime }$
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2: Proof

Whole Strategy: Step 1: see that $f^{\prime }=\varphi \circ f$, where $\varphi :{\mathbb{R}}^d\to {\mathbb{R}}^d$ is the coordinates transition map, and see that $\varphi$ is a homeomorphism; Step 2: conclude the proposition.

Step 1:

$f$ is a bijection.

$f^{\prime }$ is a bijection.

$f^{\prime }=\varphi \circ f$, where $\varphi :{\mathbb{R}}^d\to {\mathbb{R}}^d$ is the bijective coordinates transition map, where there is a constant invertible matrix, $M$, such that $\varphi :(r^1,...,r^d{)}^t\mapsto (M_j^1{r}^j,...,M_j^d{r}^j)$, 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 $C^{\mathrm{\infty }}$ manifold.

The formula for $\varphi$ means that $\varphi$ is continuous in the norm sense for the coordinates function.

So, by the proposition that for any map between $C^{\mathrm{\infty }}$ manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, $\varphi$ is continuous in the topological sense.

${\varphi }^{-1}:{\mathbb{R}}^d\to {\mathbb{R}}^d,(r^1,...,r^d)\mapsto ({M^{-1}}_j^1{r}^j,...,{M^{-1}}_j^d{r}^j)$.

So, ${\varphi }^{-1}$ is continuous, likewise.

So, $\varphi$ is a homeomorphism.

Step 2:

So, for each $U\in O$, $f^{\prime }(U)=\varphi \circ f(U)\subseteq {\mathbb{R}}^d$ is open, because $f(U)\subseteq {\mathbb{R}}^d$ is open and $\varphi$ is a homeomorphism.

That means $U\in O^{\prime }$.

By symmetry, for each $U^{\prime }\in O^{\prime }$, $U^{\prime }\in O$.

So, $O=O^{\prime }$.

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References

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