description/proof of that for Euclidean topological space, lower-dimensional Euclidean topological space, slicing map, projection, and inclusion, inclusion after projection after slicing map equals slicing map, and projection after slicing map of open neighborhood of point is open neighborhood of projection of point
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of Euclidean topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any Euclidean topological space, any lower-dimensional Euclidean topological space, the slicing map, the projection, and the inclusion, the inclusion after the projection after the slicing map equals the slicing map, and the projection after the slicing map of any open neighborhood of any point is an open neighborhood of the projection of the point.
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:
\(\mathbb{R}^{d'}\): \(= \text{ the } d' \text{ -dimensional Euclidean topological space }\)
\(J\): \(\subset \{1, ..., d'\}\) with \(\vert J \vert = d\)
\(\mathbb{R}^d\): \(= \text{ the } d \text{ -dimensional Euclidean topological space }\)
\(r'\): \(\in \mathbb{R}^{d'}\)
\(\lambda_{J, r'}\): \(: Pow (\mathbb{R}^{d'}) \to Pow (\mathbb{R}^{d'}), S \mapsto \{s \in S \vert \forall j \in \{1, ..., d'\} \setminus J (s^j = r'^j)\}\), which is "the slicing map"
\(\pi_J\): \(: \mathbb{R}^{d'} \to \mathbb{R}^d\), which takes the \(J\) components, which is "the projection"
\(\tau_{J, r'}\): \(: \mathbb{R}^d \to \mathbb{R}^{d'}\), which adds the \(\{1, ..., d'\} \setminus J\) components as those of \(r'\), which is "the inclusion"
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Statements:
\(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} = \lambda_{J, r'}\)
\(\land\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) = B_{\pi_J (r'), \epsilon})\), where \(B_{\bullet, \bullet}\) denotes the open ball with the specified center and radius
\(\land\)
\(\forall U_{r'} \in \{\text{ the open neighborhoods of } r' \text{ on } \mathbb{R}^{d'}\} (\pi_J \circ \lambda_{J, r'} (U_{r'}) \in \{\text{ the open neighborhoods of } \pi_J (r') \text{ on } \mathbb{R}^d\})\)
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2: Natural Language Description
For the \(d'\)-dimensional Euclidean topological space, \(\mathbb{R}^{d'}\), any \(J \subset \{1, ..., d'\}\) such that \(\vert J \vert = d\), the \(d\)-dimensional Euclidean topological space, \(\mathbb{R}^d\), any point, \(r' \in \mathbb{R}^{d'}\), "the slicing map", \(\lambda_{J, r'}: Pow (\mathbb{R}^{d'}) \to Pow (\mathbb{R}^{d'}), S \mapsto \{s \in S \vert \forall j \in \{1, ..., d'\} \setminus J (s^j = r'^j)\}\), "the projection", \(\pi_J: \mathbb{R}^{d'} \to \mathbb{R}^d\), which takes the \(J\) components, and "the inclusion", \(\tau_{J, r'}: \mathbb{R}^d \to \mathbb{R}^{d'}\), which adds the \(\{1, ..., d'\} \setminus J\) components as those of \(r'\), \(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} = \lambda_{J, r'}\), \(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) = B_{\pi_J (r'), \epsilon})\), where \(B_{\bullet, \bullet}\) denotes the open ball with the specified center and radius, and \(\forall U_{r'} \in \{\text{ the open neighborhoods of } r' \text{ on } \mathbb{R}^{d'}\} (\pi_J \circ \lambda_{J, r'} (U_{r'}) \in \{\text{ the open neighborhoods of } \pi_J (r') \text{ on } \mathbb{R}^d\})\).
3: Proof
Whole Strategy: Step 1: see that for each \(S \in Pow (\mathbb{R}^{d'})\), \(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (S) = \lambda_{J, r'} (S)\); Step 2: see that \(\pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) = B_{\pi_J (r'), \epsilon}\); Step 3: see that for each open neighborhood of \(r'\), \(U_{r'} \subseteq \mathbb{R}^{d'}\), \(\pi_J \circ \lambda_{J, r'} (U_{r'}) \subseteq \mathbb{R}^d\) is an open neighborhood of \(r = \pi_J (r')\).
Step 1:
Let \(S \in Pow (\mathbb{R}^{d'})\) be any.
Let \(p \in \tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (S)\) be any.
There is an \(p' \in \lambda_{J, r'} (S)\) such that \(p = \tau_{J, r'} \circ \pi_J (p')\). As the \(\{1, ..., d'\} \setminus J\) components of \(p'\) are those of \(r'\), \(p = \tau_{J, r'} \circ \pi_J (p') = p'\). So, \(p = p' \in \lambda_{J, r'} (S)\), so, \(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (S) \subseteq \lambda_{J, r'} (S)\).
Let \(p \in \lambda_{J, r'} (S)\) be any.
\(\tau_{J, r'} \circ \pi_J (p) = p\) as before. So, \(p \in \tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (S)\), so, \(\lambda_{J, r'} (S) \subseteq \tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (S)\).
So, \(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (S) = \lambda_{J, r'} (S)\).
As \(S\) is any, \(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} = \lambda_{J, r'}\).
Step 2:
Let us see that \(\pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) = B_{\pi_J (r'), \epsilon}\).
Let \(p \in \pi_J \circ \lambda_{J, r'} (B_{r', \epsilon})\) be any.
\(\tau_{J, r'} (p) \in \lambda_{J, r'} (B_{r', \epsilon})\), because while \(\tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) = \lambda_{J, r'} (B_{r', \epsilon})\), \(\tau_{J, r'} (p) \in \tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (B_{r', \epsilon})\).
So, \(\sum_{j \in J} (p^j - r'^j)^2 + \sum_{j \in \{1, ..., d'\} \setminus J} (r'^j - r'^j)^2 = \sum_{j \in J} (p^j - r'^j)^2 \lt \epsilon^2\). That means that \(p \in B_{\pi_J (r'), \epsilon}\).
So, \(\pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) \subseteq B_{\pi_J (r'), \epsilon}\).
Let \(p \in B_{\pi_J (r'), \epsilon}\) be any.
\(\sum_{j \in J} (p^j - r'^j)^2 \lt \epsilon^2\).
\(\tau_{J, r'} (p) \in B_{r', \epsilon}\), because \(\sum_{j \in J} (p^j - r'^j)^2 + \sum_{j \in \{1, ..., d'\} \setminus J} (r'^j - r'^j)^2 = \sum_{j \in J} (p^j - r'^j)^2 \lt \epsilon^2\).
\(\tau_{J, r'} (p) \in \lambda_{J, r'} (B_{r', \epsilon})\), because the \(\{1, ..., d'\} \setminus J\) components of \(\tau_{J, r'} (p)\) are those of \(r'\). So, \(\pi_J \circ \tau_{J, r'} (p) \in \pi_J \circ \lambda_{J, r'} (B_{r', \epsilon})\), but \(\pi_J \circ \tau_{J, r'} (p) = p\), and so, \(p \in \pi_J \circ \lambda_{J, r'} (B_{r', \epsilon})\).
So, \(B_{\pi_J (r'), \epsilon} \subseteq \pi_J \circ \lambda_{J, r'} (B_{r', \epsilon})\).
So, \(\pi_J \circ \lambda_{J, r'} (B_{r', \epsilon}) = B_{\pi_J (r'), \epsilon}\).
Step 3:
Let us see that for each open neighborhood of \(r'\), \(U_{r'} \subseteq \mathbb{R}^{d'}\), \(\pi_J \circ \lambda_{J, r'} (U_{r'}) \subseteq \mathbb{R}^d\) is an open neighborhood of \(r = \pi_J (r')\).
Step 3 Strategy: around each \(p \in \pi_J \circ \lambda_{J, r'} (U_{r'})\), choose an open ball contained in \(\pi_J \circ \lambda_{J, r'} (U_{r'})\), by choosing an open ball around \(p' := \tau_{J, r'} (p)\) contained in \(U_{r'}\) and taking the image of the open ball under \(\pi_J \circ \lambda_{J, p'}\).
\(r \in \pi_J \circ \lambda_{J, r'} (U_{r'})\), because \(r' \in U_{r'}\), \(r' \in \lambda_{J, r'} (U_{r'})\) because the \(\{1, ..., d'\}\) components of \(r'\) are those of \(r'\), and \(r \in \pi_J \circ \lambda_{J, r'} (U_{r'})\).
Let \(p \in \pi_J \circ \lambda_{J, r'} (U_{r'})\) be any.
\(p' := \tau_{J, r'} (p) \in \lambda_{J, r'} (U_{r'}) \subseteq U_{r'}\), because \(p' \in \tau_{J, r'} \circ \pi_J \circ \lambda_{J, r'} (U_{r'}) = \lambda_{J, r'} (U_{r'})\).
As \(U_{r'}\) is open, there is an open ball, \(B_{p', \epsilon} \subseteq \mathbb{R}^{d'}\), such that \(B_{p', \epsilon} \subseteq U_{r'}\).
\(\pi_J \circ \lambda_{J, p'} (B_{p', \epsilon}) = B_{\pi_J (p'), \epsilon} = B_{p, \epsilon}\).
\(B_{p, \epsilon} = \pi_J \circ \lambda_{J, p'} (B_{p', \epsilon}) \subseteq \pi_J \circ \lambda_{J, p'} (U_{r'})\).
But \(\pi_J \circ \lambda_{J, p'} (U_{r'}) = \pi_J \circ \lambda_{J, r'} (U_{r'})\), because the \(\{1, ..., d'\} \setminus J\) components of \(p'\) equal the \(\{1, ..., d'\} \setminus J\) components of \(r'\) and \(\lambda_{J, p'}\) depends only on the \(\{1, ..., d'\} \setminus J\) components of \(p'\).
So, \(B_{p, \epsilon} \subseteq \pi_J \circ \lambda_{J, r'} (U_{r'})\).
So, \(\pi_J \circ \lambda_{J, r'} (U_{r'})\) is an open neighborhood of \(r\).
