A description/proof of that 2 points on different connected components are not path-connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of connected topological component.
- The reader knows a definition of topological path-connected-ness of 2 points.
- The reader admits the proposition that 2 points are topologically path-connected iff there is path that connects 2 points.
- The reader admits the proposition that any continuous map image of any connected subspace of the domain is connected on the codomain.
- The reader admits the proposition that any connected subspace of any topological space is contained in a connected component.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space with multiple connected components, any 2 points on any different connected components are not 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: Description
For any topological space, \(T\), with multiple connected components, \(\{S_\alpha\}\), any 2 points, \(p_1 \in S_{\alpha'}\) and \(p_2 \in S_{\alpha''}\) where \(\alpha' \neq \alpha''\) are not path-connected on \(T\).
2: Proof
Let us suppose that \(p_1\) and \(p_2\) were path-connected on \(T\). There would be a path, \(\lambda: I \to T\), such that \(\lambda (0) = p_1\) and \(\lambda (1) = p_2\), by the proposition that 2 points are topologically path-connected iff there is path that connects 2 points. \(I\) would be connected and \(\lambda (I)\) would be connected on \(T\), by the proposition that any continuous map image of any connected subspace of the domain is connected on the codomain. As \(\lambda (I)\) would not be contained in any 1 connected component, \(\lambda (I)\) would not be connected on \(T\), by the proposition that any connected subspace of any topological space is contained in a connected component, a contradiction. So, \(p_1\) and \(p_2\) are not path-connected on \(T\).
