description/proof of that for topological space and locally finite set of closed subsets, union of set is closed
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
- 4: Proof
Starting Context
- The reader knows a definition of locally finite set of subsets of topological space.
- The reader knows a definition of closed set.
- The reader knows a definition of union of set.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any locally finite set of closed subsets, the union of the set is closed.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(A\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{C_\alpha \vert \alpha \in A\}\): \(\in \{\text{ the locally finite sets of closed subsets of } T\}\)
//
Statements:
\(\cup \{C_\alpha \vert \alpha \in A\} \subseteq T \in \{\text{ the closed subsets }\}\)
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2: Natural Language Description
For any topological space, \(T\), any possibly uncountable index set, \(A\), and any locally finite set of closed subsets of \(T\), \(\{C_\alpha \vert \alpha \in A\}\), the union of \(\{C_\alpha \vert \alpha \in A\}\), \(\cup \{C_\alpha \vert \alpha \in A\} \subseteq T\), is closed.
3: Note
When \(A\) is finite, of course, \(\cup \{C_\alpha \vert \alpha \in A\} \subseteq T\) is closed; this proposition is about a sufficient condition with which the union of some infinite number of closed subsets is closed.
4: Proof
Whole Strategy: prove that \(T \setminus \cup \{C_\alpha \vert \alpha \in A\}\) is open by the local criterion for openness; Step 1: let \(t \in T \setminus \cup \{C_\alpha \vert \alpha \in A\}\) be any and take any open neighborhood of \(t\), \(U_t \subseteq T\), that intersects only finite of \(C_\alpha\)s, \(\{C_\alpha \vert \alpha \in B\}\); Step 2: take \(U'_t := U_t \cap \cap_{\alpha \in B} (T \setminus C_\alpha) \subseteq T\) and see that \(U'_t\) is an open neighborhood of \(t\) on \(T\); Step 3: see that \(U'_t \subseteq T \setminus \cup \{C_\alpha \vert \alpha \in A\}\); Step 4: conclude the proposition.
Step 1:
Let \(t \in T \setminus \cup \{C_\alpha \vert \alpha \in A\}\) be any.
Let us take any open neighborhood of \(t\), \(U_t \subseteq T\), such that there is a finite subset (may be empty), \(B \subseteq A\), such that for each \(\alpha \in B\), \(U_t \cap C_\alpha \neq \emptyset\) and for each \(\alpha \in A \setminus B\), \(U_t \cap C_\alpha = \emptyset\), which is possible by the definition of locally finite set of subsets of topological space.
Step 2:
When \(B\) is empty, let us take \(U'_t := U_t\). \(U'_t\) is an open neighborhood of \(t\) on \(T\).
Let us suppose that \(B\) is not empty for the rest of this step.
Let us take \(U'_t := U_t \cap \cap_{\alpha \in B} (T \setminus C_\alpha) \subseteq T\).
\(t \in U'_t\), because \(t \in U_t\) and for each \(\alpha \in B\), \(t \in T \setminus C_\alpha\), because \(t \in T \setminus \cup \{C_\alpha \vert \alpha \in A\}\), which means that \(t \notin C_\alpha\) for each \(\alpha \in A\), which means that \(t \notin C_\alpha\) for each \(\alpha \in B \subseteq A\).
As \(T \setminus C_\alpha \subseteq T\) is open and \(B\) is finite, \(U'_t\) is open on \(T\) as the finite intersection of open subsets.
So, \(U'_t\) is an open neighborhood of \(t\) on \(T\).
Step 3:
Let us see that \(U'_t \subseteq T \setminus \cup \{C_\alpha \vert \alpha \in A\}\).
Let \(t' \in U'_t\) be any. \(t' \in T \setminus C_\alpha\) for each \(\alpha \in B\), which means that \(t' \notin C_\alpha\). \(t' \in U_t\), which means that \(t' \notin C_\alpha\) for each \(\alpha \in A \setminus B\) because \(U_t \cap C_\alpha = \emptyset\). So, \(t' \notin C_\alpha\) for each \(\alpha \in A\). So, \(t' \notin \cup \{C_\alpha \vert \alpha \in A\}\), which implies that \(t' \in T \setminus \cup \{C_\alpha \vert \alpha \in A\}\).
So, \(U'_t \subseteq T \setminus \cup \{C_\alpha \vert \alpha \in A\}\).
Step 4:
By the local criterion for openness, \(T \setminus \cup \{C_\alpha \vert \alpha \in A\}\) is open on \(T\), so, \(\cup \{C_\alpha \vert \alpha \in A\}\) is closed on \(T\).
