description/proof of that \(C^\infty\) manifold with boundary is locally connected
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) manifold with boundary.
- The reader knows a definition of locally connected topological space.
- The reader admits the proposition that any open set on any open topological subspace is open on the base space.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
- The reader admits the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Target Context
- The reader will have a description and a proof of the proposition that any \(C^\infty\) manifold with boundary is locally 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:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
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Statements:
\(M \in \{\text{ the locally connected topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(m \in M\) and each neighborhood of \(m\), \(N_m\), take a chart around \(m\), \((U_m \subseteq M, \phi_m)\), such that \(U_m \subseteq N_m\); Step 2: take an open ball around \(\phi_m (m)\), \(B_{\phi_m (m), \epsilon} \subseteq \phi_m (U_m)\), and see that \(\phi_m^{-1} (B_{\phi_m (m), \epsilon}) \subseteq N_m\) is a connected neighborhood of \(m\).
Step 1:
Let \(m \in M\) be any.
Let \(N_m \subseteq M\) be any neighborhood of \(m\).
There is a chart around \(m\), \((U'_m \subseteq M, \phi'_m)\).
There is an open neighborhood of \(m\), \(U''_m \subseteq M\), such that \(U''_m \subseteq N_m\), by the definition of neighborhood of point on topological space.
Let \(U_m := U'_m \cap U''_m \subseteq M\) and \(\phi_m := \phi'_m \vert_{U_m}\).
\(U_m\) is an open subset of \(M\) and \((U_m \subseteq M, \phi_m)\) is a chart around \(m\) such that \(U_m \subseteq N_m\), obviously.
Step 2:
\(\phi_m (U_m) \subseteq \mathbb{R}^d\) is an open neighborhood of \(\phi_m (m)\) on \(\mathbb{R}^d\) with the Euclidean topology.
There is an open ball around \(\phi_m (m)\), \(B_{\phi_m (m), \epsilon} \subseteq \mathbb{R}\), such that \(B_{\phi_m (m), \epsilon} \subseteq \phi_m (U_m)\), which is a property of Euclidean topology.
\(\phi_m^{-1} (B_{\phi_m (m), \epsilon}) \subseteq U_m\) is open on \(U_m\), because \(\phi_m\) is a homeomorphism.
\(\phi_m^{-1} (B_{\phi_m (m), \epsilon})\) is open on \(M\), by the proposition that any open set on any open topological subspace is open on the base space.
So, \(\phi_m^{-1} (B_{\phi_m (m), \epsilon})\) is an open neighborhood of \(m\) on \(M\).
As \(B_{\phi_m (m), \epsilon}\) is a connected subspace of \(\phi_m (U_m)\), \(\phi_m^{-1} (B_{\phi_m (m), \epsilon})\) is a connected subspace of \(U_m\), by the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
By the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, \(\phi_m^{-1} (B_{\phi_m (m), \epsilon})\) is a connected subspace of \(M\).
So, \(\phi_m^{-1} (B_{\phi_m (m), \epsilon})\) is a connected open neighborhood of \(m\) on \(M\).
\(\phi_m^{-1} (B_{\phi_m (m), \epsilon}) \subseteq U_m \subseteq N_m\).
So, \(M\) is locally connected.
