description/proof of that Euclidean topological space nested in 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 Euclidean topological space.
- The reader knows a definition of subspace topology of subset of topological space.
- The reader admits the local criterion for openness.
- The reader admits the proposition that any linear map between any Euclidean topological spaces is continuous.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting 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 \(d \le d'\)
\(\mathbb{R}^{d'}\): \(= \text{ the Euclidean topological space }\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean topological space }\)
\(\iota\): \(: \mathbb{R}^d \to \mathbb{R}^{d'}, (r^1, ..., r^d)^t \mapsto (r^1, ..., r^d, 0, ..., 0)^t\), \(= \text{ the inclusion }\)
\(r\): \(: \mathbb{R}^{d'} \to \mathbb{R}^{d'}, (r^1, ..., r^{d'})^t \mapsto M (r^1, ..., r^{d'})^t\), where \(M\) is any invertible \(d' \times d'\) real matrix
\(t\): \(: \mathbb{R}^{d'} \to \mathbb{R}^{d'}, (r^1, ..., r^{d'})^t \mapsto (r^1, ..., r^{d'})^t + (t^1, ..., t^{d'})^t\), \(= \text{ the translation }\)
\(f\): \(: \mathbb{R}^d \to t \circ r \circ \iota (\mathbb{R}^d) \subseteq \mathbb{R}^{d'}, (r^1, ..., r^d)^t \mapsto t \circ r \circ \iota ((r^1, ..., r^d)^t)\), where \(t \circ r \circ \iota (\mathbb{R}^d)\) is the topological subspace of \(\mathbb{R}^{d'}\)
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Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
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2: Note
This proposition says colloquially "\(\mathbb{R}^d\) is nested in \(\mathbb{R}^{d'}\) and \(\mathbb{R}^d\) is a topological subspace of \(\mathbb{R}^{d'}\)" but to state more exactly, \(\mathbb{R}^d\) is injected into \(\mathbb{R}^{d'}\) and the image as the topological subspace is homeomorphic to \(\mathbb{R}^d\) under the injection.
In fact, the original motivation was that \(r\) was any rotation (then, \(\mathbb{R}^d\) would be geometrically nested in \(\mathbb{R}^{d'}\)), but \(r\) does not really need to be restricted to a rotation in order for \(f\) to be homeomorphic, so, this proposition takes this generalized form.
3: Proof
Whole Strategy: Step 1: take \(\iota': \mathbb{R}^d \to \iota (\mathbb{R}^d) \subseteq \mathbb{R}^{d'}\), \(r': \iota (\mathbb{R}^d) \subseteq \mathbb{R}^{d'} \to r (\iota (\mathbb{R}^d)) \subseteq \mathbb{R}^{d'}\), and \(t': r (\iota (\mathbb{R}^d)) \subseteq \mathbb{R}^{d'} \to t (r (\iota (\mathbb{R}^d))) \subseteq \mathbb{R}^{d'}\) 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 \(\iota': \mathbb{R}^d \to \iota (\mathbb{R}^d) \subseteq \mathbb{R}^{d'}\) where the codomain is the topological subspace of \(\mathbb{R}^{d'}\), as the restriction of \(\iota\).
Let us take \(r': \iota (\mathbb{R}^d) \subseteq \mathbb{R}^{d'} \to r (\iota (\mathbb{R}^d)) \subseteq \mathbb{R}^{d'}\) where the domain and the codomain are the topological subspaces of \(\mathbb{R}^{d'}\), as the restriction of \(r\).
Let us take \(t': r (\iota (\mathbb{R}^d)) \subseteq \mathbb{R}^{d'} \to t (r (\iota (\mathbb{R}^d))) \subseteq \mathbb{R}^{d'}\) where the domain and the codomain are the topological subspaces of \(\mathbb{R}^{d'}\), 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 \(U \subseteq \mathbb{R}^d\) be any open subset.
Let us see that \(\iota' (U) \subseteq \iota (\mathbb{R}^d)\) is an open subset.
Let \(\iota (p) \in \iota' (U)\) be any.
\(p = (p^1, ..., p^d) \in U\).
There is an open ball around \(p\), \(B_{p, \epsilon} \subseteq \mathbb{R}^d\), such that \(B_{p, \epsilon} \subseteq U\).
\(\iota' (B_{p, \epsilon}) \subseteq \iota' (U)\).
Let us see that \(\iota' (B_{p, \epsilon}) = B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\), where \(B'_{\iota (p), \epsilon} \subseteq \mathbb{R}^{d'}\) is the open ball around \(\iota (p)\) on \(\mathbb{R}^{d'}\).
Let \(p' \in \iota' (B_{p, \epsilon})\) be any.
\(p' = (p'^1, ..., p'^d, 0, ..., 0)\) such that \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 \lt \epsilon^2\).
\(p' \in B'_{\iota (p), \epsilon}\), because while \(\iota (p) = (p^1, ..., p^d, 0, ..., 0)\), \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 + (0 - 0)^2 + ... + (0 - 0)^2 = (p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 \lt \epsilon^2\).
\(p' \in \iota (\mathbb{R}^d)\).
So, \(p' \in B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\).
So, \(\iota' (B_{p, \epsilon}) \subseteq B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\).
Let \(p' \in B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\) be any.
\(p' = (p'^1, ..., p'^d, 0, ..., 0)\) such that \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 + (0 - 0)^2 + ... + (0 - 0)^2 \lt \epsilon^2\).
So, \((p'^1, ..., p'^d) \in B_{p, \epsilon}\), because \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 \lt \epsilon^2\), and \(p' = \iota' ((p'^1, ..., p'^d)) \in \iota' (B_{p, \epsilon})\).
So, \(B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d) \subseteq \iota' (B_{p, \epsilon})\).
So, \(\iota' (B_{p, \epsilon}) = B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\).
So, \(\iota' (B_{p, \epsilon})\) is an open neighborhood of \(\iota (p)\) on \(\iota (\mathbb{R}^d)\), by the definition of subspace topology.
By the local criterion for openness, \(\iota' (U)\) is an open subset of \(\iota (\mathbb{R}^d)\).
Let us see the other direction.
Let \(\iota (U) \subseteq \iota (\mathbb{R}^d)\) be any open subset.
Let us see that \(U = \iota'^{-1} (\iota (U)) \subseteq \mathbb{R}^d\) is an open subset.
Let \(p \in U\) be any.
\(p = (p^1, ..., p^d) \in U\).
There is an open neighborhood of \(\iota (p)\), \(U_{\iota (p)} \subseteq \iota (\mathbb{R}^d)\), such that \(U_{\iota (p)} \subseteq \iota (U)\).
\(U_{\iota (p)} = U'_{\iota (p)} \cap \iota (\mathbb{R}^d)\), where \(U'_{\iota (p)} \subseteq \mathbb{R}^{d'}\) is an open neighborhood of \(\iota (p)\) on \(\mathbb{R}^{d'}\), by the definition of subspace topology.
There is an open ball around \(\iota (p)\), \(B'_{\iota (p), \epsilon} \subseteq \mathbb{R}^{d'}\), such that \(B'_{\iota (p), \epsilon} \subseteq U'_{\iota (p)}\).
\(B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d) \subseteq U'_{\iota (p)} \cap \iota (\mathbb{R}^d) = U_{\iota (p)} \subseteq \iota (U)\).
Let us see that \(B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d) = \iota (B_{p, \epsilon})\), where \(\iota (B_{p, \epsilon}) \subseteq \mathbb{R}^d\) is the open ball around \(p\) on \(\mathbb{R}^d\).
Let \(p' \in B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\) be any.
\(p' = (p'^1, ..., p'^d, 0, ..., 0)\) such that \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 + (0 - 0)^2 + ... + (0 - 0)^2 \lt \epsilon^2\).
So, \((p'^1, ..., p'^d) \in B_{p, \epsilon}\), because \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 \lt \epsilon^2\), and \(p' = \iota ((p'^1, ..., p'^d)) \in \iota (B_{p, \epsilon})\).
So, \(B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d) \subseteq \iota (B_{p, \epsilon})\).
Let \(p' \in \iota (B_{p, \epsilon})\) be any.
\(p' = (p'^1, ..., p'^d, 0, ..., 0)\) such that \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 \lt \epsilon^2\).
\(p' \in B'_{\iota (p), \epsilon}\), because while \(\iota (p) = (p^1, ..., p^d, 0, ..., 0)\), \((p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 + (0 - 0)^2 + ... + (0 - 0)^2 = (p'^1 - p^1)^2 + ... + (p'^d - p^d)^2 \lt \epsilon^2\).
\(p' \in \iota (\mathbb{R}^d)\).
So, \(p' \in B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\).
So, \(\iota (B_{p, \epsilon}) \subseteq B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d)\).
So, \(B'_{\iota (p), \epsilon} \cap \iota (\mathbb{R}^d) = \iota (B_{p, \epsilon})\).
So, \(\iota (B_{p, \epsilon}) \subseteq \iota (U)\).
That means that \(B_{p, \epsilon} \subseteq U\), because \(\iota\) is injective.
By the local criterion for openness, \(U\) is an open subset of \(\mathbb{R}^d\).
So, \(\iota'\) 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 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, obviously.
\(t'\) 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.
