description/proof of that complex Euclidean topological space nested in complex Euclidean topological space is topological subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of complex Euclidean topological space.
- The reader knows a definition of subspace topology of subset of topological space.
- 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 linear map between any complex Euclidean topological spaces is continuous.
Target Context
- The reader will have a description and a proof of 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.
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:
\(d'\): \(\in \mathbb{N} \setminus \{0\}\)
\(d\): \(\in \mathbb{N} \setminus \{0\}\), such that \(d \le d'\)
\(\mathbb{C}^{d'}\): \(= \text{ the complex Euclidean topological space }\)
\(\mathbb{C}^d\): \(= \text{ the complex Euclidean topological space }\)
\(\iota\): \(: \mathbb{C}^d \to \mathbb{C}^{d'}, (r^1, ..., r^d)^t \mapsto (r^1, ..., r^d, 0, ..., 0)^t\), \(= \text{ the inclusion }\)
\(r\): \(: \mathbb{C}^{d'} \to \mathbb{C}^{d'}, (r^1, ..., r^{d'})^t \mapsto M (r^1, ..., r^{d'})^t\), where \(M\) is any invertible \(d' \times d'\) complex matrix
\(t\): \(: \mathbb{C}^{d'} \to \mathbb{C}^{d'}, (r^1, ..., r^{d'})^t \mapsto (r^1, ..., r^{d'})^t + (t^1, ..., t^{d'})^t\), \(= \text{ the translation }\)
\(f\): \(: \mathbb{C}^d \to t \circ r \circ \iota (\mathbb{C}^d) \subseteq \mathbb{C}^{d'}, (r^1, ..., r^d)^t \mapsto t \circ r \circ \iota ((r^1, ..., r^d)^t)\), where \(t \circ r \circ \iota (\mathbb{C}^d)\) is the topological subspace of \(\mathbb{C}^{d'}\)
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Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
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2: Note
This proposition says colloquially "\(\mathbb{C}^d\) is nested in \(\mathbb{C}^{d'}\) and \(\mathbb{C}^d\) is a topological subspace of \(\mathbb{C}^{d'}\)" but to state more exactly, \(\mathbb{C}^d\) is injected into \(\mathbb{C}^{d'}\) and the image as the topological subspace is homeomorphic onto \(\mathbb{C}^d\) under the injection.
In fact, the word, "nested", originally came from the \(r = \text{ any rotation }\) version of the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space, and it has stuck to this version, while the situation may not be interpreted as "nested" prevalently.
3: Proof
Whole Strategy: Step 1: take \(\iota': \mathbb{C}^d \to \iota (\mathbb{C}^d) \subseteq \mathbb{C}^{d'}\), \(r': \iota (\mathbb{C}^d) \subseteq \mathbb{C}^{d'} \to r (\iota (\mathbb{C}^d)) \subseteq \mathbb{C}^{d'}\), and \(t': r (\iota (\mathbb{C}^d)) \subseteq \mathbb{C}^{d'} \to t (r (\iota (\mathbb{C}^d))) \subseteq \mathbb{C}^{d'}\) as the restrictions of \(\iota\), \(r\), and \(t\), and see that \(f = t' \circ r' \circ \iota'\); Step 2: see that \(\iota'\) is a homeomorphism; Step 3: see that \(r'\) is a homeomorphism; Step 4: see that \(t'\) is a homeomorphism; Step 5: conclude the proposition.
Step 1:
Let us take \(\iota': \mathbb{C}^d \to \iota (\mathbb{C}^d) \subseteq \mathbb{C}^{d'}\) where the codomain is the topological subspace of \(\mathbb{C}^{d'}\), as the restriction of \(\iota\).
Let us take \(r': \iota (\mathbb{C}^d) \subseteq \mathbb{C}^{d'} \to r (\iota (\mathbb{C}^d)) \subseteq \mathbb{C}^{d'}\) where the domain and the codomain are the topological subspaces of \(\mathbb{C}^{d'}\), as the restriction of \(r\).
Let us take \(t': r (\iota (\mathbb{C}^d)) \subseteq \mathbb{C}^{d'} \to t (r (\iota (\mathbb{C}^d))) \subseteq \mathbb{C}^{d'}\) where the domain and the codomain are the topological subspaces of \(\mathbb{C}^{d'}\), as the restriction of \(t\).
\(f = t' \circ r' \circ \iota'\), obviously.
As the codomain of \(\iota'\) equals the domain of \(r'\) and the codomain of \(r'\) equals the domain of \(t'\), if \(\iota'\), \(r'\), and \(t'\) are some homeomorphisms, \(f\) will be a homeomorphism, which will be proved hereafter.
Step 2:
Let us see that \(\iota'\) is a homeomorphism.
\(\iota'\) is a bijection, obviously.
Let \(f: \mathbb{R}^{2 d} \to \mathbb{C}^d\) be the canonical homeomorphism by which the topology of \(\mathbb{C}^d\) is defined.
Let \(f': \mathbb{R}^{2 d'} \to \mathbb{C}^{d'}\) be the canonical homeomorphism by which the topology of \(\mathbb{C}^{d'}\) is defined.
Let \(f'': f'^{-1} (\iota (\mathbb{C}^d)) \subseteq \mathbb{R}^{2 d'} \to \iota (\mathbb{C}^d) \subseteq \mathbb{C}^{d'}\) be the restriction of \(f'\) where the domain and the codomain are the topological subspaces of \(\mathbb{R}^{2 d'}\) and \(\mathbb{C}^{d'}\), which is a homeomorphism, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Let \(\lambda: \mathbb{R}^{2 d} \to \lambda (\mathbb{R}^{2 d}) \subseteq \mathbb{R}^{2 d'}, (r^1, ..., r^{2 d}) \mapsto (r^1, ..., r^{2 d}, 0, ..., 0)\) be the inclusion, which 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.
\(\iota' = f'' \circ \lambda \circ f^{-1}\), because for each \((r^1 + r^2 i, ..., r^{2 (d - 1) + 1} + r^{2 d} i)^t \in \mathbb{C}^d\), \(\iota' ((r^1 + r^2 i, ..., r^{2 (d - 1) + 1} + r^{2 d} i)^t ) = (r^1 + r^2 i, ..., r^{2 (d - 1) + 1} + r^{2 d} i, 0, ..., 0)^t\), while \(f'' \circ \lambda \circ f^{-1} ((r^1 + r^2 i, ..., r^{2 (d - 1) + 1} + r^{2 d} i)^t) = f'' \circ \lambda ((r^1, r^2, ..., r^{2 (d - 1) + 1}, r^{2 d})^t) = f'' ((r^1, r^2, ..., r^{2 (d - 1) + 1}, r^{2 d}, 0, ..., 0)^t) = (r^1 + r^2 i, ..., r^{2 (d - 1) + 1} + r^{2 d} i, 0, ..., 0)^t\).
The domain of \(\lambda\) equals the codomain of \(f^{-1}\) and the domain of \(f''\) equals the codomain of \(\lambda\), because \(f'^{-1} (\iota (\mathbb{C}^d)) \subseteq \mathbb{R}^{2 d'} = \lambda (\mathbb{R}^{2 d}) \subseteq \mathbb{R}^{2 d'}\).
So, \(\iota' = f'' \circ \lambda \circ f^{-1}\) is a homeomorphism.
Step 3:
Let us see that \(r'\) is a homeomorphism.
Let us see that \(r\) is a homeomorphism.
\(r\) is a linear map, so, is continuous, by the proposition that any linear map between any complex Euclidean topological spaces is continuous.
Also, \(r^{-1}\) is a linear map, because it is the map by \(M^{-1}\), so, is continuous, likewise.
\(r'\) is a homeomorphism, because \(r'\) and \(r'^{-1}\) are continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 4:
Let us see that \(t'\) is a homeomorphism.
\(t\) is a homeomorphism, because \(t = f' \circ t'' \circ f'^{-1}\) where \(t'': \mathbb{R}^{2 d'} \to \mathbb{R}^{2 d'}, (r^1, r^2, ..., r^{2 (d' - 1) + 1}, r^{2 d'})^t \mapsto (r^1, r^2, ..., r^{2 (d' - 1) + 1}, r^{2 d'})^t + (Re (t^1), Im (t^1), ..., Re (t^{d'}), Im (t^{d'}))^t\) is the translation, which is obviously a homeomorphism: for each \((r^1 + r^2 i, ..., r^{2 (d' - 1) + 1} + r^{2 d'} i)^t \in \mathbb{C}^{d'}\), \(t ((r^1 + r^2 i, ..., r^{2 (d' - 1) + 1} + r^{2 d'} i)^t) = (r^1 + r^2 i, ..., r^{2 (d' - 1) + 1} + r^{2 d'} i) + (t^1, ..., t^{d'})^t\), while \(f' \circ t'' \circ f'^{-1} ((r^1 + r^2 i, ..., r^{2 (d' - 1) + 1} + r^{2 d'} i)^t) = f' \circ t'' ((r^1, r^2, ..., r^{2 (d' - 1) + 1}, r^{2 d'})^t) = f' ((r^1, r^2, ..., r^{2 (d' - 1) + 1}, r^{2 d'})^t + (Re (t^1), Im (t^1), ..., Re (t^{d'}), Im (t^{d'}))^t) = (r^1 + r^2 i, ..., r^{2 (d' - 1) + 1} + r^{2 d'} i)^t + (t^1, ..., t^{d'})^t\).
\(t'\) is a homeomorphism, because \(t'\) and \(t'^{-1}\) are continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 5:
So, \(f\) is a homeomorphism.
