description/proof of that products of topological spaces are associative in homeomorphism sense
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of homeomorphism.
- The reader knows a definition of product topology.
- The reader admits the proposition that the nested product of any sets are associative in the 'sets - map morphisms' isomorphism sense.
Target Context
- The reader will have a description and a proof of the proposition that the nested products of any topological spaces are associative in the homeomorphism sense.
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:
\(\{T_1, T_2, T_3\}\): \(\subseteq \{\text{ the topological spaces }\}\)
\((T_1 \times T_2) \times T_3\): \(= \text{ the product topological space }\)
\(T_1 \times (T_2 \times T_3)\): \(= \text{ the product topological space }\)
\(f: (T_1 \times T_2) \times T_3 \to T_1 \times (T_2 \times T_3), ((t_1, t_2), t_3) \mapsto (t_1, (t_2, t_3))\)
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Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(f\) is a 'sets - map morphisms' isomorphism; Step 2: see that for each open subset, \(U \subseteq T_1 \times (T_2 \times T_3)\), \(f^{-1} (U) \subseteq (T_1 \times T_2) \times T_3\) is open; Step 3: see that for each open subset, \(U \subseteq (T_1 \times T_2) \times T_3\), \({f^{-1}}^{-1} (U) \subseteq T_1 \times (T_2 \times T_3)\) is open.
Step 1:
\(f\) is a 'sets - map morphisms' isomorphism, by the proposition that the nested product of any sets are associative in the 'sets - map morphisms' isomorphism sense.
So, there is the inverse, \(f^{-1}: T_1 \times (T_2 \times T_3) \to (T_1 \times T_2) \times T_3, (t_1, (t_2, t_3)) \mapsto ((t_1, t_2), t_3)\).
Step 2:
For any open subset, \(U \subseteq T_1 \times (T_2 \times T_3)\), \(U = \cup_{j \in J} (U_{1, j} \times \cup_{l \in L_j} (U_{2, j, l} \times U_{3, j, l}))\) where \(J\) and \(L_j\) are some possibly uncountable index sets and each of \(U_{m, j}\) and \(U_{m, j, l}\) is an open subset on \(T_m\), by the definition of product topology.
\(f^{-1} (U) = \cup_{j \in J} \cup_{l \in L_j} ((U_{1, j} \times U_{2, j, l}) \times U_{3, j, l})\), because for any \(t = ((t_1, t_2), t_3) \in f^{-1} (U)\), \(f (t) = (t_1, (t_2, t_3)) \in U\), \(t_1 \in U_{1, j}\) and \(t_2 \in U_{2, j, l}\) and \(t_3 \in U_{3, j, l}\) for a \(j\) and an \(l \in L_j\), so, \(t \in \cup_{j \in J} \cup_{l \in L_j} ((U_{1, j} \times U_{2, j, l}) \times U_{3, j, l})\); for any \(t = ((t_1, t_2), t_3) \in \cup_{j \in J} \cup_{l \in L_j} ((U_{1, j} \times U_{2, j, l}) \times U_{3, j, l})\), \(t_1 \in U_{1, j}\) and \(t_2 \in U_{2, j, l}\) and \(t_3 \in U_{3, j, l}\) for a \(j\) and an \(l \in L_j\), \(f (t) = (t_1, (t_2, t_3)) \in U\), so, \(t \in f^{-1} (U)\).
So, \(f^{-1} (U)\) is open on \((T_1 \times T_2) \times T_3\), by the definition of product topology.
Step 3:
For any open subset, \(U \subseteq (T_1 \times T_2) \times T_3\), \(U = \cup_{l \in L} (\cup_{j \in J_l} (U_{1, l, j} \times U_{2, l, j}) \times U_{3, l})\) where \(L\) and \(J_l\) are some possibly uncountable index sets and each of \(U_{m, l}\) and \(U_{m, l, j}\) is an open subset on \(T_m\), by the definition of product topology.
\({f^{-1}}^{-1} (U) = f (U) = \cup_{l \in L} \cup_{j \in J_l} (U_{1, l, j} \times (U_{2, l, j} \times U_{3, l}))\), because for any \(t = (t_1, (t_2, t_3)) \in {f^{-1}}^{-1} (U)\), \(f^{-1} (t) = ((t_1, t_2), t_3) \in U\), \(t_1 \in U_{1, l, j}\) and \(t_2 \in U_{2, l, j}\) and \(t_3 \in U_{3, l}\) for an \(l\) and a \(j \in J_l\), so, \(t \in \cup_{l \in L} \cup_{j \in J_l} (U_{1, l, j} \times (U_{2, l, j} \times U_{3, l}))\); for any \(t = (t_1, (t_2, t_3)) \in \cup_{l \in L} \cup_{j \in J_l} (U_{1, l, j} \times (U_{2, l, j} \times U_{3, l}))\), \(t_1 \in U_{1, l, j}\) and \(t_2 \in U_{2, l, j}\) and \(t_3 \in U_{3, l}\) for an \(l\) and a \(j \in J_l\), \(f^{-1} (t) = ((t_1, t_2), t_3) \in U\), so, \(t \in {f^{-1}}^{-1} (U)\).
So, \({f^{-1}}^{-1} (U)\) is open on \(T_1 \times (T_2 \times T_3)\), by the definition of product topology.
So, \(f\) is a homeomorphism.
3: Note
Although sloppy expressions like "\((T_1 \times T_2) \times T_3 = T_1 \times (T_2 \times T_3)\)" are prevalently seen, the 2 are not the same, but are homeomorphic to each other.
The expression, \(T_1 \times T_2 \times T_3\), is allowed because it is defined as \((T_1 \times T_2) \times T_3\), not because "\((T_1 \times T_2) \times T_3 = T_1 \times (T_2 \times T_3)\)" holds.
Base on this proposition, \(((T_1 \times T_2) \times T_3) \times T_4\) is homeomorphic to \(T_1 \times (T_2 \times (T_3 \times T_4))\) and so on, because \(((T_1 \times T_2) \times T_3) \times T_4\) is homeomorphic to \((T_1 \times (T_2 \times T_3)) \times T_4\), which is homeomorphic to \(T_1 \times ((T_2 \times T_3) \times T_4)\), which is homeomorphic to \(T_1 \times (T_2 \times (T_3 \times T_4))\).
