A description/proof of that topological 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 connected-ness of 2 points.
- The reader knows a definition of equivalence relation.
Target Context
- The reader will have a description and a proof of the proposition that topological 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, \(T\), connected-ness of 2 points is an equivalence relation.
2: Proof
For any point, \(p \in T\), \(p\) and \(p\) are connected, because \(\{p\}\) is a topological subspace that is not the union of any disjoint open sets.
For any points, \(p_1, p_2 \in T\), that are connected, \(p_2\) and \(p_1\) are connected, because there is a connected topological subspace, \(T_1\), that contains \(p_1\) and \(p_2\), which means that \(T_1\) contains \(p_2\) and \(p_1\).
For any points, \(p_1, p_2, p_3 \in T\), such that \(p_1\) and \(p_2\) are connected and \(p_2\) and \(p_3\) are connected, there are a connected topological subspace, \(T_1\), that contains \(p_1\) and \(p_2\) and a connected topological subspace, \(T_2\), that contains \(p_2\) and \(p_3\). \(T_1 \cup T_2\) is a connected topological subspace that contains \(p_1\) and \(p_3\), as is proven as follows. Suppose that \(T_1 \cup T_2\) was not connected. \(T_1 \cup T_2 = U_1 \cup U_2\), \(U_1 \cap U_2 = \emptyset\) where \(U_i\) would be a non-empty open set on \(T_1 \cup T_2\). \(U_i = {U_i}' \cap (T_1 \cup T_2)\) where \({U_i}'\) would be open on \(T\), by the definition of subspace topology. As \({U_1}' \cup {U_2}'\) would cover \(T_1 \cup T_2\), \(T_i = T_i \cap ({U_1}' \cup {U_2}') = (T_i \cap {U_1}') \cup (T_i \cap {U_2}')\). \({U_1}'\) and \({U_2}'\) would not share any point on \(T_i\), because otherwise, \(U_1\) and \(U_2\) would share the point. Each of both \(T_1\) and \(T_2\) would not be contained in a \({U_i}'\), because if both \(T_1\) and \(T_2\) were contained in \({U_1}'\) without loss of generality, \(U_2 = {U_2}' \cap (T_1 \cup T_2)\) would be empty; if \(T_1\) and \(T_2\) were contained in \({U_1}'\) and \({U_2}'\) respectively without loss of generality, \({U_1}'\) and \({U_2}'\) would share \(p_2\), then \(U_1\) and \(U_2\) would share \(p_2\). So, at least 1 of \(T_1\) and \(T_2\) would be separated into \({U_1}'\) and \({U_2}'\), which would mean that at least 1 of \(T_1\) and \(T_2\) would not be connected, a contradiction.
