description/proof of that for topological space and connected subspaces whose union is disconnected, union of nonempty subspaces of subspaces is disconnected
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.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any connected subspaces whose union is disconnected, the union of any nonempty subspaces of the subspaces is disconnected.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T'_j \subseteq T \vert j \in J\}\): \(T'_j \in \{\text{ the connected topological subspaces of } T\}\)
\(\{T_j \subseteq T'_j \vert j \in J\}\): \(T_j \in \{\text{ the topological subspaces of } T'_j\}\), such that \(T_j \neq \emptyset\)
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Statements:
\(\cup_{j \in J} T'_j \in \{\text{ the disconnected topological spaces }\}\)
\(\implies\)
\(\cup_{j \in J} T_j \in \{\text{ the disconnected topological spaces }\}\)
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2: Note
Maybe it is more important to be aware of when this proposition is not guaranteed to hold.
When a \(T'_j\) is disconnected, this proposition is not guaranteed to hold: let \(\{T'_j \subseteq T \vert j \in J\} = \{T'_1, T'_2\}\) and \(T'_1\) is disconnected, then, \(\cup_{j \in J} T'_j = U'_1 \cup U'_2\) where \(U'_1, U'_2 \subseteq \cup_{j \in J} T'_j\) are nonempty open subsets such that \(U'_1 \cap U'_2 = \emptyset\), but \(T'_1\) may be divided into \(U'_1\) and \(U'_2\) because \(T'_1\) is disconnected, and supposing \(T'_2 \subseteq U'_1\), taking \(T_1\) as the \(U'_1\) part of \(T'_1\) and taking \(T_2 = T'_2\), \(\cup_{j \in J} T_j = U'_1\) may be connected.
When a \(T_j\) is empty, this proposition is not guaranteed to hold: let \(\{T'_j \subseteq T \vert j \in J\} = \{T'_1, T'_2\}\) and \(T_1 = \emptyset\), then, taking \(T_2 = T'_2\), \(\cup_{j \in J} T_j = T'_2\) is connected.
3: Proof
Whole Strategy: Step 1: take some nonempty open subsets, \(U'_1, U'_2 \subseteq \cup_{j \in J} T'_j\), such that \(U'_1 \cup U'_2 = \cup_{j \in J} T'_j\) and \(U'_1 \cap U'_2 = \emptyset\); Step 2: see that each \(T'_j\) is in \(U'_1\) or in \(U'_2\); Step 3: see that \((\cup_{j \in J} T_j) \cap U'_1\) and \((\cup_{j \in J} T_j) \cap U'_2\) are nonempty open subsets of \(\cup_{j \in J} T_j\) such that \(((\cup_{j \in J} T_j) \cap U'_1) \cup ((\cup_{j \in J} T_j) \cap U'_2) = \cup_{j \in J} T_j\) and \(((\cup_{j \in J} T_j) \cap U'_1) \cap ((\cup_{j \in J} T_j) \cap U'_2) = \emptyset\).
Step 1:
There are some nonempty open subsets, \(U'_1, U'_2 \subseteq \cup_{j \in J} T'_j\), such that \(U'_1 \cup U'_2 = \cup_{j \in J} T'_j\) and \(U'_1 \cap U'_2 = \emptyset\), by the definition of connected topological space.
That means that there are some open subsets, \(U_1, U_2 \subseteq T\), such that \(U'_1 = U_1 \cap \cup_{j \in J} T'_j\) and \(U'_2 = U_2 \cap \cup_{j \in J} T'_j\), by the definition of subspace topology.
Step 2:
Each \(T'_j\) is in \(U'_1\) or in \(U'_2\), because otherwise, \(T'_j = (T'_j \cap U_1) \cup (T'_j \cap U_2)\) (because \(T'_j = (T'_j \cap U'_1) \cup (T'_j \cap U'_2) \subseteq (T'_j \cap U_1) \cup (T'_j \cap U_2) \subseteq T'_j\)), where \(T'_j \cap U_1\) and \(T'_j \cap U_2\) would be some open subsets of \(T'_j\), and \((T'_j \cap U_1) \cap (T'_j \cap U_2) = \emptyset\), because if \(t \in (T'_j \cap U_1) \cap (T'_j \cap U_2)\), \(t \in U_1 \cap \cup_{j \in J} T'_j = U'_1\) and \(t \in U_2 \cap \cup_{j \in J} T'_j = U'_2\), so, \(t \in U'_1 \cap U'_2\), a contradiction, and \(T'_j \cap U_1\) and \(T'_j \cap U_2\) would be nonempty, a contradiction against \(T'_j\)'s being connected.
A \(T'_j\) is in \(U'_1\), because otherwise, \(U'_1\) would be empty, a contradiction; likewise, a \(T'_j\) is in \(U'_2\).
Step 3:
\((\cup_{j \in J} T_j) \cap U'_1 = (\cup_{j \in J} T_j) \cap (U_1 \cap \cup_{j \in J} T'_j) = (\cup_{j \in J} T_j) \cap \cup_{j \in J} T'_j \cap U_1 = (\cup_{j \in J} T_j) \cap U_1\) is an open subset of \(\cup_{j \in J} T_j\).
It is nonempty, because a \(T'_j\) is in \(U'_1\) and so, the nonempty \(T_j\) is in \(U'_1\).
Likewise, \((\cup_{j \in J} T_j) \cap U'_2\) is a nonempty open subset of \(\cup_{j \in J} T_j\).
\(((\cup_{j \in J} T_j) \cap U'_1) \cup ((\cup_{j \in J} T_j) \cap U'_2) = \cup_{j \in J} T_j\), because for each \(t \in \cup_{j \in J} T_j\), \(t \in U'_1\) or \(t \in U'_2\).
\(((\cup_{j \in J} T_j) \cap U'_1) \cap ((\cup_{j \in J} T_j) \cap U'_2) = \emptyset\), because \(U'_1 \cap U'_2 = \emptyset\).
So, \(\cup_{j \in J} T_j\) is disconnected.
