description/proof of that for topological space and set of connected subspaces, if each pair of subspaces is not disjoint, union of subspaces is connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of connected topological space.
- The reader knows a definition of subspace topology of subset of topological space.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any set of connected subspaces, if each pair of the subspaces is not disjoint, the union of the subspaces is 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the connected topological subspaces of } T\} \vert j \in J\}\):
//
Statements:
\(\forall j_1, j_2 \in J (T_{j_1} \cap T_{j_2}) \neq \emptyset\)
\(\implies\)
\(\cup_{j \in J} T_j \in \{\text{ the connected topological subspaces of } T\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(\cup_{j \in J} T_j\) was not connected, and find a contradiction.
Step 1:
Let us suppose that \(\cup_{j \in J} T_j\) was not connected.
There would be some nonempty open subsets, \(U_1, U_2 \subseteq \cup_{j \in J} T_j\), such that \(\cup_{j \in J} T_j = U_1 \cup U_2\) and \(U_1 \cap U_2 = \emptyset\).
\(U_1 = U'_1 \cap (\cup_{j \in J} T_j)\) and \(U_2 = U'_2 \cap (\cup_{j \in J} T_j)\) where \(U'_1, U'_2 \subseteq T\) were some open subsets of \(T\), by the definition of subspace topology.
For each \(j \in J\), \(T_j \subseteq \cup_{j \in J} T_j = U_1 \cup U_2 \subseteq U'_1 \cup U'_2\).
So, \(T_j = T_j \cap (U'_1 \cup U'_2) = (T_j \cap U'_1) \cup (T_j \cap U'_2)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
But \((T_j \cap U'_1) \cap (T_j \cap U'_2) \subseteq (U'_1 \cap (\cup_{j \in J} T_j)) \cap (U'_2 \cap (\cup_{j \in J} T_j)) = U_1 \cap U_2 = \emptyset\), where \(T_j \cap U'_1\) and \(T_j \cap U'_2\) were some open subsets of \(T_j\).
As \(T_j\) was connected, \(T_j \cap U'_2 = \emptyset\) or \(T_j \cap U'_1 = \emptyset\), so, \(T_j = T_j \cap U'_1\) or \(T_j = T_j \cap U'_2\), accordingly.
There would be a \(j_1 \in J\) such that \(T_{j_1} = T_{j_1} \cap U'_1\), because otherwise, \(U_1 = U'_1 \cap (\cup_{j \in J} T_j) = \cup_{j \in J} (U'_1 \cap T_j)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(= \cup_{j \in J} \emptyset = \emptyset\), a contradiction.
Likewise, there would be a \(j_2 \in J\) such that \(T_{j_2} = T_{j_2} \cap U'_2\).
By the supposition, there would be a \(t \in T_{j_1} \cap T_{j_2} = (T_{j_1} \cap U'_1) \cap (T_{j_2} \cap U'_2)\), but \(T_{j_1} \cap U'_1 \subseteq U'_1 \cap (\cup_{j \in J} T_j) = U_1\) and \(T_{j_2} \cap U'_2 \subseteq U'_2 \cap (\cup_{j \in J} T_j) = U_2\), so, \(t \in (T_{j_1} \cap U'_1) \cap (T_{j_2} \cap U'_2) \subseteq U_1 \cap U_2\), a contradiction against \(U_1 \cap U_2 = \emptyset\).
So, \(\cup_{j \in J} T_j\) is connected.
