description/proof of that for product topological space and partition of index set, product space is homeomorphic to product of subproduct topological spaces with divided index sets
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 homeomorphism.
- The reader admits the proposition that for any product set and any set of subsets of each constituent set, the product of the unions of the subsets of the constituent sets is the union of the products of the subsets.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
- The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
- The reader admits the proposition that any open continuous bijection is a homeomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any product topological space and any partition of the index set, the product space is homeomorphic to the product of the subproduct topological spaces with the divided index sets.
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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(L\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{J_l \subseteq J \vert l \in L\}\): such that for each \(l_1, l_2 \in L\) such that \(l_1 \neq l_2\), \(J_{l_1} \cap J_{l_2} = \emptyset\) and \(\cup_{l \in L} J_l = J\)
\(\times_{l \in L} \times_{j \in J_l} T_j\): \(= \text{ the product topological space }\)
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Statements:
\(\times_{j \in J} T_j \equiv_{homeomorphism} \times_{l \in L} \times_{j \in J_l} T_j\)
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2: Proof
Whole Strategy: Step 1: define \(f: \times_{j \in J} T_j \to \times_{l \in L} \times_{j \in J_l} T_j, \times_{j \in J} t_j \mapsto \times_{l \in L} \times_{j \in J_l} t_j\) and see that \(f\) is a bijection; Step 2: see that \(f\) is continuous; Step 3: see that \(f\) is open; Step 4: conclude the proposition.
Step 1:
Let us define \(f: \times_{j \in J} T_j \to \times_{l \in L} \times_{j \in J_l} T_j, \times_{j \in J} t_j \mapsto \times_{l \in L} \times_{j \in J_l} t_j\).
Let us see that \(f\) is a bijection.
Let \(\times_{j \in J} t_j, \times_{j \in J} t'_j \in \times_{j \in J} T_j\) be any such that \(\times_{j \in J} t_j \neq \times_{j \in J} t'_j\).
There is a \(j \in J\) such that \(t_j \neq t'_j\).
There is a \(l \in L\) such that \(j \in J_l\).
\(\times_{j \in J_l} t_j \neq \times_{j \in J_l} t'_j\).
So, \(\times_{l \in L} \times_{j \in J_l} t_j \neq \times_{l \in L} \times_{j \in J_l} t'_j\).
So, \(f\) is injective.
Let \(\times_{l \in L} \times_{j \in J_l} t_j \in \times_{l \in L} \times_{j \in J_l} T_j\) be any.
There is \(t \in \times_{j \in J} T_j\) such that the \(j\)-component is \(t^j = t_j\).
Then, \(t = \times_{j \in J} t_j\), and \(f (t) = \times_{l \in L} \times_{j \in J_l} t_j\).
So, \(f\) is surjective.
So, \(f\) is bijective.
Step 2:
Let \(U \subseteq \times_{l \in L} \times_{j \in J_l} T_j\) be any open subset.
\(U = \cup_{m \in M} \times_{l \in L} U_{m, l}\), where \(M\) is a possibly uncountable index set and \(U_{m, l} \subseteq \times_{j \in J_l} T_j\) is an open subset such that for each \(m \in M\), only finite of \(U_{m, l}\) s are not \(\times_{j \in J_l} T_j\) s, by Note for the definition of product topology.
For each \(m\), for each \(l \in L\) such that \(U_{m, l} = \times_{j \in J_l} T_j\), \(U_{m, l} = \cup_{n_{m, l} \in N_{m, l}} \times_{j \in J_l} U_{m, l, n_{m, l}, j}\), where \(\vert N_{m, l} \vert = 1\) and \(U_{m, l, n_{m, l}, j} = T_j\); for each \(l \in L\) such that \(U_{m, l} \neq \times_{j \in J_l} T_j\), \(U_{m, l} = \cup_{n_{m, l} \in N_{m, l}} \times_{j \in J_l} U_{m, l, n_{m, l}, j}\), where \(N_{m, l}\) is a possibly uncountable index set and \(U_{m, l, n_{m, l}, j} \subseteq T_j\) is an open subset such that for each \(n_{m, l} \in N_{m, l}\), only finite of \(U_{m, l, n_{m, l}, j}\) s are not \(T_j\) s, by Note for the definition of product topology.
So, \(U = \cup_{m \in M} \times_{l \in L} \cup_{n_{m, l} \in N_{m, l}} \times_{j \in J_l} U_{m, l, n_{m, l}, j}\).
By the proposition that for any product set and any set of subsets of each constituent set, the product of the unions of the subsets of the constituent sets is the union of the products of the subsets, \(= \cup_{m \in M} \cup_{g \in \times_{l \in L} N_{m, l}} \times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}\): take \(S_{l, n_{m, l}} := \times_{j \in J_l} U_{m, l, n_{m, l}, j}\), then \(\times_{l \in L} \cup_{n_{m, l} \in N_{m, l}} \times_{j \in J_l} U_{m, l, n_{m, l}, j} = \times_{l \in L} \cup_{n_{m, l} \in N_{m, l}} S_{l, n_{m, l}}\), and \(= \cup_{g \in \times_{l \in L} N_{m, l}} \times_{l \in L} S_{l, g (l)} = \cup_{g \in \times_{l \in L} N_{m, l}} \times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}\).
\(f^{-1} (U) = f^{-1} (\cup_{m \in M} \cup_{g \in \times_{l \in L} N_{m, l}} \times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}) = \cup_{m \in M} \cup_{g \in \times_{l \in L} N_{m, l}} f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j})\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
Let us see that \(f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}) = \times_{j \in J} U_{m, l, g (l), j}\).
Let \(f^{-1} (\times_{l \in L} \times_{j \in J_l} u_{m, l, g (l), j}) \in f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j})\) be any.
\(f (\times_{j \in J} u_{m, l, g (l), j}) = \times_{l \in L} \times_{j \in J_l} u_{m, l, g (l), j}\), so, \(f^{-1} (\times_{l \in L} \times_{j \in J_l} u_{m, l, g (l), j}) = \times_{j \in J} u_{m, l, g (l), j} \in \times_{j \in J} U_{m, l, g (l), j}\).
So, \(f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}) \subseteq \times_{j \in J} U_{m, l, g (l), j}\).
Let \(\times_{j \in J} u_{m, l, g (l), j} \in \times_{j \in J} U_{m, l, g (l), j}\) be any.
\(f (\times_{j \in J} u_{m, l, g (l), j}) = \times_{l \in L} \times_{j \in J_l} u_{m, l, g (l), j} \in \times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}\), so, \(\times_{j \in J} u_{m, l, g (l), j} \in f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j})\).
So, \(\times_{j \in J} U_{m, l, g (l), j} \subseteq f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j})\).
So, \(f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j}) = \times_{j \in J} U_{m, l, g (l), j}\).
Only finite of \(U_{m, l, g (l), j}\) s are not \(T_j\) s, because while \(g \in \times_{l \in L} N_{m, l}\), there is a finite \(L^` \subseteq L\) such that for each \(l \in L \setminus L^`\), \(\vert N_{m, l} \vert = 1\) and \(U_{m, l, g (l), j} = T_j\) and for each \(l \in L^`\), only some finite of \(U_{m, l, g (l), j}\) s are not \(T_j\) s.
So, \(f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j})\) is open on \(\times_{j \in J} T_j\).
So, \(f^{-1} (U) = \cup_{m \in M} \cup_{g \in \times_{l \in L} N_{m, l}} f^{-1} (\times_{l \in L} \times_{j \in J_l} U_{m, l, g (l), j})\) is open on \(\times_{j \in J} T_j\).
So, \(f\) is continuous.
Step 3:
Let \(U \subseteq \times_{j \in J} T_j\) be any open subset.
\(U = \cup_{m \in M} \times_{j \in J} U_{m, j}\), where \(M\) is a possibly uncountable index set and \(U_{m, j} \subseteq T_j\) is an open subset such that for each \(m \in M\), only finite of \(U_{m, j}\) s are not \(T_j\) s, as before.
\(f (U) = f (\cup_{m \in M} \times_{j \in J} U_{m, j}) = \cup_{m \in M} f (\times_{j \in J} U_{m, j})\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
Let us see that \(f (\times_{j \in J} U_{m, j}) = \times_{l \in L} \times_{j \in J_l} U_{m, j}\).
Let \(f (\times_{j \in J} u_{m, j}) \in f (\times_{j \in J} U_{m, j})\) be any.
\(f (\times_{j \in J} u_{m, j}) = \times_{l \in L} \times_{j \in J_l} u_{m, j} \in \times_{l \in L} \times_{j \in J_l} U_{m, j}\).
So, \(f (\times_{j \in J} U_{m, j}) \subseteq \times_{l \in L} \times_{j \in J_l} U_{m, j}\).
Let \(\times_{l \in L} \times_{j \in J_l} u_{m, j} \in \times_{l \in L} \times_{j \in J_l} U_{m, j}\) be any.
\(\times_{j \in J} u_{m, j} \in \times_{j \in J} U_{m, j}\) and \(f (\times_{j \in J} u_{m, j}) = \times_{l \in L} \times_{j \in J_l} u_{m, l}\), so, \(\times_{l \in L} \times_{j \in J_l} u_{m, l} \in f (\times_{j \in J} U_{m, j})\).
So, \(\times_{l \in L} \times_{j \in J_l} U_{m, j} \subseteq f (\times_{j \in J} U_{m, j})\).
So, \(f (\times_{j \in J} U_{m, j}) = \times_{l \in L} \times_{j \in J_l} U_{m, j}\).
As only finite of \(U_{m, j}\) s are not \(T_j\) s, for each \(l \in L\), \(\times_{j \in J_l} U_{m, j}\) is open on \(\times_{j \in J_l} T_j\), and only finite of \(\times_{j \in J_l} U_{m, j}\) s are not \(\times_{j \in J_l} T_j\) s, so, \(\times_{l \in L} \times_{j \in J_l} U_{m, j}\) is open on \(\times_{l \in L} \times_{j \in J_l} T_j\).
So, \(f (U) = \cup_{m \in M} \times_{l \in L} \times_{j \in J_l} U_{m, j}\) is open on \(\times_{l \in L} \times_{j \in J_l} T_j\).
So, \(f\) is open.
Step 4:
So, \(f\) is a homeomorphism, by the proposition that any open continuous bijection is a homeomorphism.
