description/proof of that finite product of topological sums is topological sum of products
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 topological sum.
- 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.
- The reader admits the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
Target Context
- The reader will have a description and a proof of the proposition that any finite product of any topological sums is the topological sum of the products of the constituents of the topological sums.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{L_j \in \{\text{ the index sets }\} \vert j \in J\}\):
\(\{T_{j, l_j} \in \{\text{ the topological spaces }\} \vert j \in J, l_j \in L_j\}\)
\(T\): \(= \times_{j \in J} \coprod_{l_j \in L_j} T_{j, l_j}\)
\(T'\): \(= \coprod_{\times_{j \in J} l_j \in \times_{j \in J} L_j} \times_{j \in J} T_{j, l_j}\)
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Statements:
\(T = T'\)
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2: Proof
Whole Strategy: Step 1: see that \(T\) and \(T'\) are the same set-wise; Step 2: see that each open \(U \subseteq T\) is open on \(T'\); Step 3: see that each open \(U' \subseteq T'\) is open on \(T\).
Step 1:
Let \(J = \{1, ..., n\}\) without loss of generality, just for our convenience of expressions.
\(T\) and \(T'\) are the same set-wise, because there is the identity map, \(id: T \to T', (p_1, ..., p_n) \mapsto (p_1, ..., p_n)\), where \((p_1, ..., p_n) \in T\) means that \(p_1 \in T_{1, l_1}, ..., p_n \in T_{n, l_n}\), so, \((p_1, ..., p_n) \in \times_{j \in J} T_{j, l_j}\) in \(T'\), while it is indeed an injection, because for any \((p_1, ..., p_n) \neq (p'_1, ..., p'_n)\), \(p_j \neq p'_j\) for a \(j \in J\), and \(id ((p_1, ..., p_n))^j = p_j \neq p'_j = id ((p'_1, ..., p'_n))^j\), and is a surjection, because for each \((p_1, ..., p_n) \in T'\), \(id ((p_1, ..., p_n)) = (p_1, ..., p_n)\).
The issue is whether \(T\) and \(T'\) have the same topology.
Step 2:
Let \(U \subseteq T\) be any open subset.
\(U = \cup_{m \in M} \times_{j \in j} U_{j, m}\), where \(M\) is a possibly uncountable index set and \(U_{j, m} \subseteq \coprod_{l_j \in L_j} T_{j, l_j}\) is open on \(\coprod_{l_j \in L_j} T_{j, l_j}\), by Note for the definition of product topology.
\(U \cap \times_{j \in J} T_{j, l_j} = (\cup_{m \in M} \times_{j \in j} U_{j, m}) \cap (\times_{j \in J} T_{j, l_j}) = \cup_{m \in M} (\times_{j \in j} U_{j, m} \cap \times_{j \in J} T_{j, l_j})\), 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, \(= \cup_{m \in M} \times_{j \in J} (U_{j, m} \cap T_{j, l_j})\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
For each \(j \in J\), \(U_{j, m} \cap T_{j, l_j}\) is open on \(T_{j, l_j}\), because \(U_{j, m}\) is open on \(\coprod_{l_j \in L_j} T_{j, l_j}\), by the definition of topological sum.
So, \(\times_{j \in J} (U_{j, m} \cap T_{j, l_j})\) is open on \(\times_{j \in J} T_{j, l_j}\), by the definition of product topology, so, \(U \cap \times_{j \in J} T_{j, l_j}\) is open on \(\times_{j \in J} T_{j, l_j}\).
So, \(U\) is open on \(T'\), by the definition of topological sum.
Step 3:
Let \(U' \subseteq T'\) be any open subset.
\(U' \cap (\times_{j \in J} T_{j, l_j})\) is open on \(\times_{j \in J} T_{j, l_j}\), by the definition of topological sum.
So, \(U' \cap (\times_{j \in J} T_{j, l_j}) = \cup_{m \in M_{\times_{j \in J} l_j}} \times_{j \in J} U_{j, m}\), where \(M_{\times_{j \in J} l_j}\) is a possibly uncountable index set and \(U_{j, m} \subseteq T_{j, l_j}\) is open on \(T_{j, l_j}\), by Note for the definition of product topology.
\(U' = \cup_{\times_{j \in J} l_j \in \times_{j \in J} L_j} (U' \cap (\times_{j \in J} T_{j, l_j})) = \cup_{\times_{j \in J} l_j \in \times_{j \in J} L_j} \cup_{m \in M_{\times_{j \in J} l_j}} \times_{j \in J} U_{j, m}\).
But \(U_{j, m}\) is open on \(\coprod_{l_j \in L_j} T_{j, l_j}\), because \(U_{j, m} \cap T_{j, l_j} = U_{j, m}\) is open on \(T_{j, l_j}\) and \(U_{j, m} \cap T_{j, l'_j} = \emptyset\) for any \(l'_j \neq l_j\), by the definition of topological sum.
So, \(\times_{j \in J} U_{j, m}\) is open on \(T\), by the definition of product topology, and \(U' = \cup_{\times_{j \in J} l_j \in \times_{j \in J} L_j} \cup_{m \in M_{\times_{j \in J} l_j}} \times_{j \in J} U_{j, m}\) is open on \(T\).
