A description/proof of that topological path-connected-ness of 2 points is equivalence relation
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of topological path-connected-ness of 2 points.
- The reader knows a definition of equivalence relation.
- The reader admits the proposition that any 2 points are path-connected on any topological space if and only if there is a path that connects the 2 points on the topological space.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
Target Context
- The reader will have a description and a proof of the proposition that topological path-connected-ness of 2 points is an equivalence relation.
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, \(T\), path-connected-ness of 2 points is an equivalence relation.
2: Proof
By the proposition that any 2 points are path-connected on any topological space if and only if there is a path that connects the 2 points on the topological space, we only need to cite a path on \(T\) in order to prove that 2 concerned points are path-connected.
For any point, \(p \in T\), \(p\) and \(p\) are path-connected, because the curve, \(\lambda: [0, 1] \rightarrow T\), which constantly maps to \(p\) is a path.
For any points, \(p_1, p_2 \in T\), that are path-connected, \(p_2\) and \(p_1\) are path-connected, because there is a path, \(\lambda: [0, 1] \rightarrow T\), where \(\lambda (0) = p_1\) and \(\lambda (1) = p_2\), so, there is the path, \({\lambda}': [0, 1] \rightarrow [0, 1] \rightarrow T\), where its 1st-half, \([0, 1] \rightarrow [0, 1]\), is \(r \mapsto 1 - r\) and its 2nd-half, \([0, 1] \rightarrow T\), is \(\lambda\), continuous as a compound of continuous maps, with \({\lambda}' (0) = p_2\) and \({\lambda}' (1) = p_1\).
For any points, \(p_1, p_2, p_3 \in T\), such that \(p_1\) and \(p_2\) are path-connected and \(p_2\) and \(p_3\) are path-connected, there are paths, \({\lambda}_1: [0, 2^{-1}] \rightarrow T\) where \({\lambda}_1 (0) = p_1\) and \({\lambda}_1 (2^{-1}) = p_2\) and \({\lambda}_2: [2^{-1}, 1] \rightarrow T\) where \({\lambda}_2 (2^{-1}) = p_2\) and \({\lambda}_2 (1) = p_3\). Let us define \({\lambda}': [0, 1] \rightarrow T\) as \({\lambda}'|_{[0, 2^{-1}]} = {\lambda}_1\) and \(\lambda'|_{[2^{-1}, 1]} = {\lambda}_2\). \({\lambda}'\) is continuous by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous, and \({\lambda}' (0) = p_1\) and \({\lambda}' (1) = p_3\). So, \(p_1\) and \(p_3\) are path-connected.
