A description/proof of that connected topological component is exactly connected topological subspace that cannot be made larger
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of connected topological component.
Target Context
- The reader will have a description and a proof of the proposition that any connected topological component is exactly any 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 connected topological component, \(T_1 \subseteq T\), is exactly any connected topological subspace that cannot be made larger.
2: Proof
Is \(T_1\) connected? Suppose that \(T_1\) was not connected. \(T_1 = U_1 \cup U_2\), \(U_1 \cap U_2 = \emptyset\) where \(U_i\) would be a non-empty open set on \(T_1\). \(U_i = {U_i}' \cap T_1\) where \({U_i}'\) would be an open set on \(T\), by the definition of subspace topology. There would be 2 points, \(p_1, p_2 \in T_1\), such that \(p_1 \in {U_1}'\) and \(p_2 \in {U_2}'\). As \(p_1\) and \(p_2\) would be connected, there would be a connected subspace, \(p_1, p_2 \in T_2 \subseteq T\), but in fact, \(T_2 \subseteq T_1\), because all the points on \(T_2\) belong to the equivalence class. \(T_2 = T_2 \cap ({U_1}' \cup {U_2}') = (T_2 \cap {U_1}') \cup (T_2 \cap {U_2}')\), because \(U_1 \cup U_2\) would contain \(T_1\), so, \({U_1}' \cup {U_2}'\) would contain \(T_2\). \({U_1}'\) and \({U_2}'\) would not share any point on \(T_2\), because otherwise, \(U_1\) and \(U_2\) would share the point. So, \(T_2\) would not be connected, a contradiction, so, \(T_1\) is connected.
Adding any point to \(T_1\) makes the result not a connected topological subspace, because the added point does not belong to the equivalence class, which means that there is no 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 connected.
Suppose that \(T_3\) is any 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 connected subspace, \(T_1 \subseteq T_3\), so, \(T_3 = T_1\).
3: Note
It is not so obvious that any connected topological component is connected, because connected topological component is defined based on connected-ness of any pair of points in the component, which is about the existence of a connected topological subspace, which is not necessarily the component; the component is certainly the union of such connected subspaces, but there is no guarantee that such a union is connected, although the union of any sequence of connected subspaces that share a point pair-wise sequentially is connected.
