A description/proof of that finite product of topological spaces equals sequential products of topological spaces
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of product of sets.
Target Context
- The reader will have a description and a proof of the proposition that the product of any finite number of topological spaces equals the sequential products of the topological spaces.
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: Description
For any topological spaces, \(T_1, T_2, . . ., T_n\), the product topological space, \(T_1 \times T_2 \times . . . \times T_n\), equals the sequential products, \(( . . . (T_1 \times T_2) . . . ) \times T_n\).
2: Proof
Sets-product-wise, \(T_1 \times T_2 \times . . . \times T_n\) is nothing but \(( . . . (T_1 \times T_2) . . . ) \times T_n\).
1st, let us think of the case, \(n = 3\). Any open set, \(U\), on \(T_1 \times T_2 \times T_3\) is \(U = \cup_\alpha (U_{1-\alpha} \times U_{2-\alpha} \times U_{3-\alpha})\), by the definition of product topology. Any open set, \(U'\), on \((T_1 \times T_2) \times T_n\) is \(U' = \cup_\beta (\cup_{\gamma} (U_{1-\beta-\gamma} \times U_{2-\beta-\gamma}) \times U_{3-\beta})\), by the definition of product topology.
In order to realize \(U\) by \(U'\), \(\{i-\beta-\gamma\}\) can be taken to be 1 element for each \(\beta\), then, \(U' = \cup_\beta ((U_{1-\beta-1} \times U_{2-\beta-1}) \times U_{3-\beta}) = \cup_\beta (U_{1-\beta-1} \times U_{2-\beta-1} \times U_{3-\beta})\), and \(\{U_{i-\beta-1}\}\) and \(\{U_{3-\beta}\}\) can be taken to be \(\{U_{i-\alpha}\}\).
In order to realize \(U'\) by \(U\), some \(U_{3-\alpha}\)s can be chosen to be the same for some multiple \(\alpha\)s, then, the new indexes set, \(\{\beta-\gamma\}\), can be chosen such that \(\{U_{i-\beta-\gamma}\} = \{U_{i-\alpha}\}\) where \(U_{3-\beta-\gamma}\)s are the same for the same \(\beta\), then, \(U = \cup_\alpha (U_{1-\alpha} \times U_{2-\alpha} \times U_{3-\alpha}) = \cup_{\beta-\gamma} (U_{1-\beta-\gamma} \times U_{2-\beta-\gamma} \times U_{3-\beta-\gamma}) = \cup_{\beta} (\cup_{\gamma} (U_{1-\beta-\gamma} \times U_{2-\beta-\gamma}) \times U_{3-\beta})\) where \(U_{3-\beta-\gamma}\) is written as \(U_{3-\beta}\) as it does not really depend on \(\gamma\).
So, the 2 topologies are the same.
Obviously, the case for any \(n\) can be proven likewise.
