description/proof of that for topological space, closed subset, and disjoint set of open subsets whose union contains closed subset, intersection of closed subset and each open subset is closed on space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closed subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, any closed subset, and any disjoint set of open subsets whose union contains the closed subset, the intersection of the closed subset and each open subset is closed on the space.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(C\): \(\in \{\text{ the closed subsets of } T\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{U_j \vert j \in J\}\): \(U_j \in \{\text{ the open subsets of } T\}\) such that for each \(j, j' \in J\) such that \(j \neq j'\), \(U_j \cap U_{j'} = \emptyset\), and \(C \subseteq \cup_{j \in J} U_j\)
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Statements:
\(\forall j \in J (C \cap U_j \in \{\text{ the closed subsets of } T\})\)
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2: Proof
Whole Strategy: Step 1: see that \(C \cap U_{j'} = C \cap (T \setminus \cup_{j \in J \setminus \{j'\}} U_j)\), and conclude the proposition.
Step 1:
Let us see that \(C \cap U_{j'} = C \cap (T \setminus \cup_{j \in J \setminus \{j'\}} U_j)\).
For each \(c \in C \cap U_{j'}\), \(c \in T\), and for each \(j \in J \setminus \{j'\}\), \(c \notin U_j\), because \(U_j \cap U_{j'} = \emptyset\), so, \(c \notin \cup_{j \in J \setminus \{j'\}} U_j\), so, \(c \in C \cap (T \setminus \cup_{j \in J \setminus \{j'\}} U_j)\).
For each \(c \in C \cap (T \setminus \cup_{j \in J \setminus \{j'\}} U_j)\), \(c \notin \cup_{j \in J \setminus \{j'\}} U_j\), so, for each \(j \in J \setminus \{j'\}\), \(c \notin U_j\), so, \(c \in U_{j'}\), because \(c \in \cup_{j \in J} U_j\), so, \(c \in C \cap U_{j'}\).
So, \(C \cap U_{j'} = C \cap (T \setminus \cup_{j \in J \setminus \{j'\}} U_j)\).
As \(\cup_{j \in J \setminus \{j'\}} U_j\) is open on \(T\), \(T \setminus \cup_{j \in J \setminus \{j'\}} U_j\) is closed on \(T\), and \(C \cap (T \setminus \cup_{j \in J \setminus \{j'\}} U_j)\) is a closed subset as an intersection of closed subsets.
So, \(C \cap U_{j'}\) is closed on \(T\).
