description/proof of that for finite-dimensional real or complex vectors space with canonical topology, canonical topology of vectors subspace is subspace topology
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 knows a definition of subspace topology of subset of topological space.
- The reader admits the proposition that for any vectors space and any linearly independent subset, the subset can be expanded to be a basis.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space.
- The reader admits the proposition that any complex Euclidean topological space nested in any complex Euclidean topological space is a topological subspace of the nesting complex Euclidean topological space.
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, the canonical topology of any vectors subspace is the subspace topology.
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
\(d'\): \(\in \mathbb{N} \setminus \{0\}\)
\(d\): \(\in \mathbb{N} \setminus \{0\}\), such that \(d \le d'\)
\(V'\): \(\in \{\text{ the } d' \text{ -dimensional } F \text{ vectors spaces }\}\), with the canonical topology
\(V\): \(\in \{\text{ the } d \text{ -dimensional vectors subspaces of } V'\}\)
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Statements:
\(\text{ the canonical topology of } V = \text{ the subspace topology of } V \text{ in } V'\)
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2: Note
As the canonical topology and the subspace topology have been defined differently, we need a proof for the claim.
3: Proof
Whole Strategy: Step 1: take the inclusion, \(\iota: V \to \iota (V) \subseteq V'\), and see that the proposition equals that \(\iota\) is a homeomorphism; Step 2: take any basis for \(V\), \(B\), and any basis for \(V'\), \(B'\), as an expansion of \(B\), and take the canonical homeomorphism, \(f: V \to F^d\), and the canonical homeomorphism, \(f': V' \to F^{d'}\), with respect to the bases, and take the restriction of \(f'\), \(f'': \iota (V) \to f' (\iota (V))\); Step 3: take the inclusion, \(\iota': F^d \to \iota' (F^d) \subseteq F^{d'}\), and see that \(\iota = f''^{-1} \circ \iota' \circ f\); Step 4: conclude the proposition.
Step 1:
Let us take the inclusion, \(\iota: V \to \iota (V) \subseteq V'\), where \(V\) has the canonical topology and \(\iota (V)\) has the subspace topology.
The proposition equals that \(\iota\) is a homeomorphism, because if \(\iota\) is homeomorphic, for each open subset of \(V\), \(U\), \(\iota (U)\) is an open subset of \(\iota (V)\), and for each open subset of \(\iota (V)\), \(\iota (U)\), \(U\) is an open subset of \(V\), which equals that \(V\) and \(\iota (V)\) have the same topology.
Step 2:
Let us take any basis for \(V\), \(B = \{b_1, ... b_d\}\).
\(B\) is linearly independent on \(V'\), because otherwise, \(B\) would not be linearly independent on \(V\).
By the proposition that for any vectors space and any linearly independent subset, the subset can be expanded to be a basis, a basis, \(B'\), can be chosen such that \(B \subseteq B'\), so, \(B' = \{b_1, ..., b_d, b'_{d + 1}, ..., b'_{d'}\}\).
Let us take the canonical homeomorphism, \(f: V \to F^d\), with respect to \(B\).
Let us take the canonical homeomorphism, \(f': V' \to F^{d'}\), with respect to \(B'\).
Let us take the restriction of \(f'\), \(f'': \iota (V) \subseteq V' \to f' (\iota (V)) \subseteq F^{d'}\), where \(\iota (V)\) has the subspace topology of \(V'\) and \(f' (\iota (V))\) has the subspace topology of \(F^{d'}\).
\(f''\) is a homeomorphism, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 3:
Let us take the inclusion, \(\iota': F^d \to \iota' (F^d) \subseteq F^{d'}, (r^1, ..., r^d) \mapsto (r^1, ..., r^d, 0, ..., 0)\) where \(\iota' (F^d)\) has the subspace topology of \(F^{d'}\).
\(\iota'\) is a homeomorphism, by the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space or the proposition that any complex Euclidean topological space nested in any complex Euclidean topological space is a topological subspace of the nesting complex Euclidean topological space.
\(\iota = f''^{-1} \circ \iota' \circ f\), because for each \(v = v^j b_j \in V\), \(\iota (v) = v = v^j b_j\), while \(f''^{-1} \circ \iota' \circ f (v) = f''^{-1} \circ \iota' (v^1, ..., v^d) = f''^{-1} (v^1, ..., v^d, 0, ..., 0) = v^j b_j + \sum_{j \in \{d + 1, ..., d'\}} 0 b'_j = v^j b_j\).
Step 4:
The domain of \(\iota'\) equals the codomain of \(f\) and the domain of \(f''^{-1}\) equals the codomain of \(\iota'\), because \(f' (\iota (V)) \subseteq F^{d'} = \iota' (F^d) \subseteq F^{d'}\).
So, \(\iota\) is a homeomorphism.
