A description/proof of that path-connected topological component is exactly path-connected topological subspace that cannot be made larger
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that any path-connected topological component is exactly any path-connected topological subspace that cannot be made larger.
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 topological space, \(T\), any path-connected topological component, \(T_1 \subseteq T\), is exactly any path-connected topological subspace that cannot be made larger.
2: Proof
Is \(T_1\) path-connected? For any points, \(p_1, p_2 \in T_1\), as \(p_1\) and \(p_2\) are path-connected on \(T\), there is a path, \(\lambda: [0, 1] \rightarrow T\). \(\lambda ([0, 1]) \subseteq T_1\), because for any point, \(p_3 \in \lambda ([0, 1])\), \(p_3 = \lambda (r_3)\), and by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \({\lambda}': [0, r_3] \rightarrow T = {\lambda}|_{[0, r_3]}\) is a path on \(T\) that connects \(p_1\) and \(p_3\), so, \(p_3 \in T_1\). By the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \({\lambda}'': [0, 1] \rightarrow T_1\) as the restriction of \(\lambda\), is a path on \(T_1\). So, yes, \(T_1\) is path-connected.
Adding any point to \(T_1\) makes the result not a path-connected topological subspace, because the added point does not belong to the equivalence class, which means that there is no path-connected topological subspace that contains the added point and a point of \(T_1\), so, the result subspace that contains the both points cannot be path-connected.
Suppose that \(T_3\) is any path-connected topological subspace that contains a point of \(T_1\) and cannot be made larger. All the points of \(T_3\) belong to the equivalence class of the point, so, \(T_3 \subseteq T_1\), but as \(T_1\) is a path-connected subspace, \(T_1 \subseteq T_3\), so, \(T_3 = T_1\).
3: Note
It is not so obvious that any path-connected topological component is path-connected, because path-connected topological component is defined based on path-connected-ness of any pair of points on the component, which is about the existence of a path-connected topological subspace, which is not necessarily the component; the component is certainly the union of such path-connected subspaces, but there is no guarantee that such a union is path-connected, although the union of any sequence of path-connected subspaces that share a point pair-wise sequentially is path-connected.
