A description/proof of that path-connected topological component is open and closed on locally path-connected topological space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of path-connected topological component.
- The reader knows a definition of locally path-connected topological space.
- The reader knows a definition of closed set.
- The reader admits the proposition that any 2 points that are path-connected on any topological subspace are path-connected on any larger subspace.
- The reader admits the local criterion for openness.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that any path-connected topological component is open and closed on any locally path-connected 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: Description
For any locally path-connected topological space, \(T\), any path-connected component, \(T_1 \subseteq T\), is open and closed.
2: Proof
Around any point, \(p \in T_1\), there are a neighborhood, \(p \in N_p \subseteq T\), and a path-connected neighborhood, \(p \in {N_p}' \subseteq N_p\). \({N_p}' \subseteq T_1\), because each point on \({N_p}'\) is path-connected with \(p\) on \({N_p}'\), so, also on \(T\), by the proposition that any 2 points that are path-connected on any topological subspace are path-connected on any larger subspace. So, \(T_1\) is open by the local criterion for openness.
Suppose that \(T_1\) was not closed. While the closure \(\overline{T_1}\) would be closed, there would be a point, \(p \in \overline{T_1}\) and \(p \notin T_1\). By the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, \(p\) would be an accumulation point of \(T_1\). For any neighborhood of \(p\), \(N_p\), there would be a path-connected neighborhood, \(p \in {N_p}' \subseteq N_p\). As \(p\) would be an accumulation point of \(T_1\), \({N_p}'\) would contain a point of \(T_1\), so, \(p\) would be path-connected with the point of \(T_1\) on \({N_p}'\), and would be so on \(T\), by the proposition that any 2 points that are path-connected on any topological subspace are path-connected on any larger subspace. So, \(p\) would be in \(T_1\), a contradiction. So, \(T_1\) is closed.
