description/proof of that Euclidean topological space is homeomorphic to product of lower-dimensional Euclidean spaces
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: Note
- 4: Proof
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of product topology.
Target Context
- The reader will have a description and a proof of the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\).
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:
\(\mathbb{R}^d\): \(= \text{ the Euclidean topological space }\), where \(d \in \mathbb{N} \setminus \{0\}\)
\((\mathbb{R}^{d_1}, ..., \mathbb{R}^{d_n})\): \(\in \{\text{ the sequences of Euclidean topological spaces }\}\), where \(d_j \in \mathbb{N} \setminus \{0\}\), such that \(\sum_{j \in \{1, ..., n\}} d_j = d\)
//
Statements:
\(\mathbb{R}^d \cong \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\), where \(\cong\) denotes being homeomorphic.
//
2: Natural Language Description
For the \(d\)-dimensional Euclidean topological space, \(\mathbb{R}^d\), where \(d \in \mathbb{N} \setminus \{0\}\), and any sequence of Euclidean topological spaces, \((\mathbb{R}^{d_1}, ..., \mathbb{R}^{d_n})\), where \(d_j \in \mathbb{N} \setminus \{0\}\), such that \(\sum_{j \in \{1, ..., n\}} d_j = d\), \(\mathbb{R}^d\) is homeomorphic to \(\mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\).
3: Note
Prevalently, \(\mathbb{R}^d\) will be sloppily said to equal \(\mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\), but they are not exactly the same sets-wise. For example, a point on \(\mathbb{R}^3\) is \((r_1, r_2, r_3)\) while a point on \(\mathbb{R}^1 \times \mathbb{R}^2\) is \((r_1, (r_2, r_3))\), and the 2 points are different, strictly speaking. As they are not the same sets-wise, they cannot be the same 'topological spaces'-wise. So, this proposition says that the 2 topological spaces are homeomorphic, not the same.
4: Proof
Whole Strategy: Step 1: define the canonical 'sets - map morphisms' isomorphism, \(f: \mathbb{R}^d \to \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\); Step 2: see that \(f\) is continuous at each point, \(p \in \mathbb{R}^d\); Step 3: see that \(f^{-1}\) is continuous at each point, \(q \in \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\).
Step 1:
Let us denote any point, \(p \in \mathbb{R}^d\), as \(p = (p^1, ..., p^d)\).
Let us denote any point, \(q \in \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\), as \(q = (q^1, ..., q^n)\).
There is the canonical 'sets - map morphisms' isomorphism, \(f: \mathbb{R}^d \to \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}, (p^1, ..., p^d) \mapsto (q^1 = (p^1, ..., p^{d_1}), ..., q^n = (p^{d_1 + ... + d_{n - 1} + 1}, ..., p^d))\), which is indeed bijective, because for each \(p, p' \in \mathbb{R}^d\) such that \(p \neq p'\), \(p^j \neq p'^j\) for a \(j \in \{1, ..., d\}\), and there is a \(k \in \{1, ..., n\}\) such that \(p^j\) is in \(q^k\) and \(p'^j\) is in \(q'^k\), which means that \(q^k \neq q'^k\), which means that \(f (p) \neq f (p')\); for each \(q = (q^1 = (p^1, ..., p^{d_1}), ..., q^n = (p^{d_1 + ... + d_{n - 1} + 1}, ..., p^d)) \in \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\), there is the \(p = (p^1, ..., p^d) \in \mathbb{R}^d\), which means that \(f (p) = q\).
Step 2:
Let us see that \(f\) is continuous at any point, \(p = (p^1, ..., p^d) \in \mathbb{R}^d\).
Let \(q := f (p) = (q^1, ..., q^n)\).
Let \(N_q \subseteq \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\) be any neighborhood of \(q\). There is an open neighborhood of \(q\), \(U_q \subseteq \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\), such that \(U_q \subseteq N_q\). There is an open \(B_{q^1, \epsilon_1} \times ... \times B_{q^n, \epsilon_n} \subseteq \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\) such that \(B_{q^1, \epsilon_1} \times ... \times B_{q^n, \epsilon_1} \subseteq U_q\), where \(B_{q^j, \epsilon_j} \subseteq \mathbb{R}^{d_j}\) is the open ball around \(q^j\), by the definition of product topology.
Let \(\epsilon := min (\epsilon_1, ..., \epsilon_n)\).
Let us take the open ball around \(p\), \(B_{p, \epsilon} \subseteq \mathbb{R}^d\).
Let us see that \(f (B_{p, \epsilon}) \subseteq B_{q^1, \epsilon_1} \times ... \times B_{q^n, \epsilon_n}\).
Let \(p' = (p'^1, ..., p'^d) \in B_{p, \epsilon}\) be any. \(f (p') = (q'^1 = (p'^1, ..., p'^{d_1}), ..., q'^n = (p'^{d_1 + ... + d_{n - 1} + 1}, ..., p'^{d}))\). \(q'^j \in B_{q^j, \epsilon_j}\), because \(\sum_{j \in \{d_1 + ... + d_{j - 1} + 1, ..., d_1 + ... + d_{j - 1} + d_j\}} (p'^j - p^j)^2 \le \sum_{j \in \{1, ..., d\}} (p'^j - p^j)^2 \lt \epsilon^2 \le \epsilon_j^2\). So, \(f (p') \in B_{q^1, \epsilon_1} \times ... \times B_{q^n, \epsilon_n}\). So, \(f (B_{p, \epsilon}) \subseteq B_{q^1, \epsilon_1} \times ... \times B_{q^n, \epsilon_n}\).
So, \(f (B_{p, \epsilon}) \subseteq B_{q^1, \epsilon_1} \times ... \times B_{q^n, \epsilon_n} \subseteq U_q \subseteq N_q\).
So, \(f\) is continuous at \(p\).
Step 3:
Let us see that \(f^{-1}\) is continuous at any point, \(q = (q^1 = (p^1, ..., p^{d_1}), ..., q^n = (p^{d_1 + ... + d_{n - 1} + 1}, ..., p^d)) \in \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\).
Let \(p := f^{^1} (q) = (p^1, ..., p^n)\).
Let \(N_p \subseteq \mathbb{R}^d\) be any neighborhood of \(p\). There is an open neighborhood of \(p\), \(U_p \subseteq \mathbb{R}^d\), such that \(U_p \subseteq N_p\). There is an open \(B_{p, \epsilon} \subseteq \mathbb{R}^d\) such that \(B_{p, \epsilon} \subseteq U_p\), where \(B_{p, \epsilon} \subseteq \mathbb{R}^d\) is the open ball around \(p\).
Let \(\epsilon' := \epsilon / \sqrt{n}\).
Let us take the open ball around \(q\), \(B_{q^1, \epsilon'} \times ... \times B_{q^n, \epsilon'} \subseteq \mathbb{R}^{d_1} \times . . . \times \mathbb{R}^{d_n}\).
Let us see that \(f^{-1} (B_{q^1, \epsilon'} \times ... \times B_{q^n, \epsilon'}) \subseteq B_{p, \epsilon}\).
Let \(q' = (q'^1 = (p'^1, ..., p'^{d_1}), ..., q'^n = (p'^{d_1 + ... + d_{n - 1} + 1}, ..., p'^d)) \in B_{q^1, \epsilon'} \times ... \times B_{q^n, \epsilon'}\) be any. \(p' := f^{-1} (q') = (p'^1, ..., p'^d)\). \(p' \in B_{p, \epsilon}\), because \(\sum_{j \in \{1, ..., d\}} (p'^j - p^j)^2 = \sum_{j \in \{1, ..., d_1\}} (p'^j - p^j)^2 + ...+ \sum_{j \in \{d_1 + ... + d_{n - 1} + 1, ..., d\}} (p'^j - p^j)^2 \lt \epsilon'^2 + ...+ \epsilon'^2 = n \epsilon^2 / n = \epsilon^2\). So, \(f^{-1} (q') \in B_{p, \epsilon}\). So, \(f^{-1} (B_{q^1, \epsilon'} \times ... \times B_{q^n, \epsilon'}) \subseteq B_{p, \epsilon}\).
So, \(f^{-1} (B_{q^1, \epsilon'} \times ... \times B_{q^n, \epsilon'}) \subseteq B_{p, \epsilon} \subseteq U_p \subseteq N_p\).
So, \(f^{-1}\) is continuous at \(q\).
