A description/proof of that product of closed sets 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 topology.
- The reader knows a definition of closed set.
- The reader admits the proposition that the product of any complements is the product of the whole sets minus the union of the products of the whole sets, 1 of which is replaced with the subset for each constituent set.
Target Context
- The reader will have a description and a proof of the proposition that any product of closed sets (1 from each constituent topological space) is closed in the product topology.
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 product topological space, \(T := T_1 \times T_2 \times . . . \times T_n\), and any closed sets, \(C_i \subseteq T_i\), \(C_1 \times C_2 \times . . . \times C_n \subseteq T\) is closed on \(T\).
2: Proof
\(C_1 \times C_2 \times . . . \times C_n = (T_1 \setminus U_1) \times (T_2 \setminus U_2) \times . . . \times (T_n \setminus U_n)\) where \(U_i \subseteq T_i\) is open on \(T_i\). \((T_1 \setminus U_1) \times (T_2 \setminus U_2) \times . . . \times (T_n \setminus U_n) = T_1 \times T_2 \times . . . \times T_n \setminus ((U_1 \times T_2 \times . . . \times T_n) \cup (T_1 \times U_2 \times . . . \times T_n) \cup . . . \cup (T_1 \times T_2 \times . . . \times U_n))\), by the proposition that the product of any complements is the product of the whole sets minus the union of the products of the whole sets, 1 of which is replaced with the subset for each constituent set. As each \(T_1 \times T_2 \times . . . \times U_i \times . . . \times T_n\) is open, \(((U_1 \times T_2 \times . . . \times T_n) \cup (T_1 \times U_2 \times . . . \times T_n) \cup . . . \cup (T_1 \times T_2 \times . . . \times U_n))\) is open.
