description/proof of that for finite-dimensional real or complex vectors space with canonical topology and open subset, subset of open subset s.t. each element has its inverse contained in open subset is open subset
Topics
About: vectors space
About: topological 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 knows a definition of canonical topology for finite-dimensional complex vectors space.
- The reader admits the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional real or complex vectors space with the canonical topology and any open subset, the subset of the open subset such that each element has its inverse contained in the open subset is an open subset.
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
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with the canonical topology
\(U\): \(\in \{\text{ the open subsets of } V\}\)
\(S\): \(= \{v \in U \vert - v \in U\}\)
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Statements:
\(S \in \{\text{ the open subsets of } V\}\)
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2: Note
Inevitably, for each \(v \in S\), \(- v \in S\), because \(- (- v) = v \in U\).
\(S\) may be empty.
3: Proof
Whole Strategy: Step 1: take the canonical homeomorphism, \(f: V \to \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\), and see that \(f\) is a 'real vectors spaces - real linear morphisms' isomorphism; Step 2: for each \(v \in S\), take an open ball around \(v\), \(B_{v, \epsilon}\), and an open ball around \(-v\), \(B_{- v, \epsilon}\), contained in \(U\), and see that \(B_{v, \epsilon} \subseteq S\).
Step 1:
There is the canonical homeomorphism with respect to any basis, \(B = \{b_1, ..., b_d\}\), \(f': V \to F^d\), by the definition of canonical topology for finite-dimensional real vectors space or the definition of canonical topology for finite-dimensional complex vectors space.
When \(F = \mathbb{R}\), let us take \(f = f': V \to \mathbb{R}^d\).
When \(F = \mathbb{C}\), there is the canonical homeomorphism, \(f'': F^d \to \mathbb{R}^{2 d}\), and \(f = f'' \circ f': V \to \mathbb{R}^{2 d}\) is a homeomorphism.
Let us see that \(f\) is a 'real vectors spaces - real linear morphisms' isomorphism.
Let us see that \(f\) is real linear.
When \(F = \mathbb{R}\), for each \(v, v' \in V\) and each \(r, r' \in \mathbb{R}\), \(f (r v + r' v') = f (r v^j b_j + r' v'^j b_j) = f ((r v^j + r' v'^j) b_j) = (r v^1 + r' v'^1, ..., r v^d + r' v'^d) = r (v^1, ..., v^d) + r' (v'^1, ..., v'^d) = r f (v) + r' f (v')\).
When \(F = \mathbb{C}\), for each \(v, v' \in V\) and each \(r, r' \in \mathbb{R}\), \(f (r v + r' v') = f (r v^j b_j + r' v'^j b_j) = f ((r v^j + r' v'^j) b_j) = f'' (r v^1 + r' v'^1, ..., r v^d + r' v'^d) = f'' (r Re (v^1) + r' Re (v'^1) + (r Im (v^1) + r' Im (v'^1)) i, ..., r Re (v^d) + r' Re (v'^d) + (r Im (v^d) + r' Im (v'^d)) i) = (r Re (v^1) + r' Re (v'^1), r Im (v^1) + r' Im (v'^1), ..., r Re (v^d) + r' Re (v'^d), r Im (v^d) + r' Im (v'^d)) = r (Re (v^1), Im (v^1), ..., Re (v^d), Im (v^d)) + r' (Re (v'^1), Im (v'^1), ..., Re (v'^d), Im (v'^d)) = r f'' (v^1, ..., v^d) + r' f'' (v'^1, ..., v'^d) = r f (v) + r' f (v')\).
\(f\) is bijective.
So, \(f\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
Step 2:
Let \(v \in S\) be any.
When we talk of the open ball around \(v\), \(B_{v, \epsilon} \subseteq V\), that means \(f^{-1} (B_{f (v), \epsilon})\) where \(B_{f (v), \epsilon} \subseteq \mathbb{R}^{d} \text{ or } \mathbb{R}^{2 d}\) is the open ball around \(f (v)\).
As \(v \in U\) and \(U\) is open, there is an open ball around \(v\), \(B_{v, \epsilon'} \subseteq V\), such that \(B_{v, \epsilon'} \subseteq U\): \(f (U) \subseteq \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\) is open, there is an open ball around \(f (v)\), \(B_{f (v), \epsilon'} \subseteq \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\), such that \(B_{f (v), \epsilon'} \subseteq f (U)\), and \(B_{v, \epsilon'} = f^{-1} (B_{f (v), \epsilon'}) \subseteq U\).
As \(- v \in U\) and \(U\) is open, there is an open ball around \(- v\), \(B_{- v, \epsilon''} \subseteq V\), such that \(B_{- v, \epsilon''} \subseteq U\): \(f (U) \subseteq \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\) is open, there is an open ball around \(f (- v)\), \(B_{f (- v), \epsilon''} \subseteq \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\), such that \(B_{f (- v), \epsilon''} \subseteq f (U)\), and \(B_{- v, \epsilon''} = f^{-1} (B_{f (- v), \epsilon''}) \subseteq U\).
Let \(\epsilon := min (\epsilon', \epsilon'')\).
Then, \(B_{v, \epsilon} \subseteq U\) and \(B_{- v, \epsilon} \subseteq U\).
Let us see that \(B_{v, \epsilon} \subseteq S\).
Let us see that \(- B_{v, \epsilon} = B_{- v, \epsilon}\).
For each \(v' \in - B_{v, \epsilon}\), \(v' = - v''\) for a \(v'' \in B_{v, \epsilon}\), so, \(\Vert f (v'') - f (v) \Vert \lt \epsilon\), but \(\Vert f (v'') - f (v) \Vert = \Vert - f (v'') + f (v) \Vert = \Vert f (- v'') - f (- v) \Vert\), because \(f\) is linear, \(= \Vert f (v') - f (- v) \Vert \lt \epsilon\), which means that \(v' \in B_{- v, \epsilon}\).
So, \(- B_{v, \epsilon} \subseteq B_{- v, \epsilon}\).
For each \(v' \in B_{- v, \epsilon}\), \(\Vert f (v') - f (- v) \Vert \lt \epsilon\), but \(\Vert f (v') - f (- v) \Vert = \Vert - f (v') + f (- v) \Vert = \Vert f (- v') - f (v) \Vert\), because \(f\) is linear, \(\lt \epsilon\), which means that \(- v' \in B_{v, \epsilon}\), which means that \(v' = - (- v') \in - B_{v, \epsilon}\).
So, \(B_{- v, \epsilon} \subseteq - B_{v, \epsilon}\).
So, \(- B_{v, \epsilon} = B_{- v, \epsilon}\).
So, for each \(v \in B_{v, \epsilon} \subseteq U\), \(- v \in - B_{v, \epsilon} = B_{- v, \epsilon} \subseteq U\), which means that \(v \in S\).
So, \(B_{v, \epsilon} \subseteq S\).
As \(B_{v, \epsilon} = f^{-1} (B_{f (v), \epsilon})\) is open on \(V\), \(S\) is open, by the local criterion for openness.
