A description/proof of that union of 2 connected subspaces is connected if each neighborhood of point on subspace contains point of other subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of subspace topology.
- The reader knows a definition of connected topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace.
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, \(T\), and any connected subspaces, \(T_1, T_2 \subseteq T\), \(T_1 \cup T_2\) is connected if there is a point, \(p \in T_1\), whose each neighborhood, \(U_p \subseteq T\), contains a point on \(T_2\).
2: Proof
Suppose that there is such a point, \(p\). 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 nonempty open on \(T_1 \cup T_2\), and \(U_i = {U_i}' \cap (T_1 \cup T_2)\) where \({U_i}'\) would be open on \(T\). Without loss of generality, \(p \in U_1\). \({U_1}'\) would be a neighborhood of \(p\), so, would contain a point, \(p' \in T_2\). \(T_2 = (({U_1}' \cup {U_2}') \cap T_2)\), as \(U_1 \cup U_2\) would contain \(T_2\) and \({U_1}' \cup {U_2}'\) would contain \(U_1 \cup U_2\). \(T_2 = ({U_1}' \cap T_2) \cup ({U_2}' \cap T_2)\). \({U_1}'\) and \({U_2}'\) would not share any point on \(T_2\), because otherwise, \(U_1\) and \(U_2\) would share the point. As \(T_2\) would be connected, \({U_2}' \cap T_2\) would have to be empty as \(p' \in {U_1}'\). So, \({U_2}'\) would have to contain a point of \(T_1\), but then, \(T_1 = ({U_1}' \cup {U_2}') \cap T_1 = ({U_1}' \cap T_1) \cup ({U_2}' \cap T_1)\), a union of disjoint nonempty open sets, because \({U_1}'\) and \({U_2}'\) would not share any point on \(T_1\), a contradiction. So, \(T_1 \cup T_2\) is connected.
