description/proof of that connected topological manifold is path-connected
Topics
About: topological manifold
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of topological manifold.
- The reader knows a definition of connected topological space.
- The reader knows a definition of path-connected topological space.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that any connected topological manifold is path-connected.
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:
\(M\): \(\in \{\text{ the connected topological manifolds }\}\)
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Statements:
\(M\): \(\in \{\text{ the path-connected topological manifolds }\}\)
//
2: Natural Language Description
Any connected topological manifold, \(M\), is path-connected.
3: Proof
Whole Strategy: take any point, \(p \in M\), and the subset, \(S \subseteq M\), as all the points path-connected with \(p\), and see that \(S = M\); Step 1: take any point, \(p \in M\), and the subset, \(S \subseteq M\), as all the points path-connected with \(p\); Step 2: see that \(S\) is open; Step 3: see that \(M \setminus S\) is open; Step 4: conclude the proposition.
Step 1:
For any point, \(p \in M\), take the subset, \(S \subseteq M\), as all the points path-connected with \(p\).
Step 2:
Let us prove that \(S\) is open.
Let \(p_1 \in S\) be any. There are an open neighborhood of \(p_1\), \(U_{p_1} \subseteq M\), and a homeomorphism, \(\phi_{p_1}: U_{p_1} \to U_{\phi_{p_1} (p_1)}\), where \(U_{\phi_{p_1} (p_1)} \subseteq \mathbb {R}^n\) is open on \(\mathbb {R}^n\).
There is an open ball, \(B_{\phi_{p_1} (p_1), \epsilon} \subseteq \mathbb {R}^n\), such that \(B_{\phi_{p_1} (p_1), \epsilon} \subseteq U_{\phi_{p_1} (p_1)}\). \(B_{\phi_{p_1} (p_1), \epsilon}\) is open on \(U_{\phi_{p_1} (p_1)}\), and \(\phi_{p_1}^{-1} (B_{\phi_{p_1} (p_1), \epsilon}) \subseteq U_{p_1}\) is open on \(U_{p_1}\) and on \(M\).
Any \(p_2 \in \phi_{p_1}^{-1} (B_{\phi_{p_1} (p_1), \epsilon})\) is path-connected to \(p_1\) on \(\phi_{p_1}^{-1} (B_{\phi_{p_1} (p_1), \epsilon})\) and on \(M\), because \(\phi_{p_1} (p_2)\) is path-connected to \(\phi_{p_1} (p_1)\) on \(B_{\phi_{p_1} (p_1), \epsilon}\) and the path can be mapped into \(\phi_{p_1}^{-1} (B_{\phi_{p_1} (p_1), \epsilon})\) by \(\phi_{p_1}^{-1}\): for the path, \(\lambda: J \subseteq \mathbb{R} \to B_{\phi_{p_1} (p_1), \epsilon}\), take \(\phi_{p_1}^{-1} \circ \lambda: J \to \phi_{p_1}^{-1} (B_{\phi_{p_1} (p_1), \epsilon})\), which is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(p_2\) is path-connected to \(p\) on \(M\), because \(p_2\) can be path-connected to \(p\) via \(p_1\). So, \(\phi_{p_1}^{-1} (B_{\phi_{p_1} (p_1), \epsilon}) \subseteq S\).
By the local criterion for openness, \(S\) is open on \(M\).
Step 3:
Let us prove that \(M \setminus S \subseteq M\) is open.
Let \(p_3 \in M \setminus S\) be any. There are an open neighborhood of \(p_3\), \(U_{p_3} \subseteq M\), and a homeomorphism, \(\phi_{p_3}: U_{p_3} \to U_{\phi (p_3)}\), where \(U_{\phi_{p_3} (p_3)} \subseteq \mathbb {R}^n\) is open on \(\mathbb {R}^n\). There is an open ball, \(B_{\phi_{p_3} (p_3), \epsilon} \subseteq \mathbb {R}^n\), such that \(B_{\phi_{p_3} (p_3), \epsilon} \subseteq U_{\phi_{p_3} (p_3)}\). \(B_{\phi_{p_3} (p_3), \epsilon}\) is open on \(U_{\phi_{p_3} (p_3)}\), and \(\phi_{p_3}^{-1} (B_{\phi_{p_3} (p_3), \epsilon}) \subseteq U_{p_3}\) is open on \(U_{p_3}\) and on \(M\).
Any \(p_4 \in \phi_{p_3}^{-1} (B_{\phi_{p_3} (p_3), \epsilon})\) is path-connected to \(p_3\) on \(\phi_{p_3}^{-1} (B_{\phi_{p_3} (p_3), \epsilon})\) and on \(M\), because \(\phi_{p_3} (p_4)\) is path-connected to \(\phi_{p_3} (p_3)\) on \(B_{\phi_{p_3} (p_3), \epsilon}\) and the path can be mapped into \(\phi_{p_3}^{-1} (B_{\phi_{p_3} (p_3), \epsilon})\) by \(\phi_{p_3}^{-1}\), as before.
\(p_4\) is not path-connected to \(p\) on \(M\), because otherwise, \(p_3\) would be path-connected to \(p\) via \(p_4\), a contradiction. So, \(\phi_{p_3}^{-1} (B_{\phi_{p_3} (p_3), \epsilon}) \subseteq M \setminus S\).
By the local criterion for openness, \(M \setminus S\) is open on \(M\).
Step 4:
Now \(S \cap (M \setminus S) = \emptyset\) and \(M = S \cup (M \setminus S)\). As \(M\) is connected, one of the 2 subsets has to be empty, but \(S\) is not empty containing at least \(p\), so, \(M \setminus S = \emptyset\), and so, \(S = M\), which means that \(M\) is path-connected.
4: Note
This proposition is not about topological space in general, but about topological manifold.
