description/proof of that product of closed subsets is closed in product topology
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topological space.
- The reader knows a definition of closed subset of topological space.
- The reader admits the proposition that for any product set, the product of the complements of any subsets is the product of the whole sets minus the union of the products of the whole sets, \(1\) of which is replaced with a subset for each product.
Target Context
- The reader will have a description and a proof of the proposition that for any product topological space, the product of any closed subsets 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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(\{C_j \in \{\text{ the closed subsets of } T_j\} \vert j \in J\}\):
//
Statements:
\(\times_{j \in J} C_j \in \{\text{ the closed subsets of } \times_{j \in J} T_j\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(C_j = T_j \setminus U_j\) and \(\times_{j \in J} C_j = \times_{j \in J} (T_j \setminus U_j) = (\times_{j \in J} T_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\) where \(S^`_{j', j'} = U_{j'}\) and \(S^`_{j', j} = T_j\) for each \(j \in J \setminus \{j'\}\); Step 2: conclude the proposition.
Step 1:
For each \(j \in J\), \(C_j = T_j \setminus U_j\) where \(U_j \subseteq T_j\) is an open subset.
\(\times_{j \in J} C_j = \times_{j \in J} (T_j \setminus U_j) = (\times_{j \in J} T_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\) where \(S^`_{j', j'} = U_{j'}\) and \(S^`_{j', j} = T_j\) for each \(j \in J \setminus \{j'\}\), by the proposition that for any product set, the product of the complements of any subsets is the product of the whole sets minus the union of the products of the whole sets, \(1\) of which is replaced with a subset for each product.
Step 2:
For each \(j' \in J\), \(\times_{j \in J} S^`_{j', j} \subseteq \times_{j \in J} T_j\) is open, because each \(S^`_{j', j}\) is open and only finite (in fact, 1) of \(S^`_{j', j}\) s are not \(T_j\) s, by Note for the definition of product topology.
\(\cup_{j' \in J} \times_{j \in J} S^`_{j', j}\) is open as a union of open subsets.
So, \((\times_{j \in J} T_j) \setminus (\cup_{j' \in J} \times_{j \in J} S^`_{j', j})\) is closed.
So, \(\times_{j \in J} C_j\) is closed.
