description/proof of that projection of product topological space is open
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 open map.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any projection of any product topological space is open.
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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(l\): \(\in J\)
\(\pi^l\): \(: \times_{j \in J} T_j \to T_l, f \mapsto f (l)\), \(= \text{ the projection }\)
//
Statements:
\(\pi^l \in \{\text{ the open maps }\}\)
//
2: Note
Compare with the proposition that a projection of a product topological space is not necessarily closed.
3: Proof
Whole Strategy: Step 1: see that for each open \(U \subseteq \times_{j \in J} T_j\), \(U = \cup_{j' \in J'} \times_{j \in J} U_{j', j}\); Step 2: see that \(\pi^l (U) = \cup_{j' \in J'} U_{j', l}\).
Step 1:
Let \(U \subseteq \times_{j \in J} T_j\) be any open subset.
\(U = \cup_{j' \in J'} \times_{j \in J} U_{j', j}\) where \(J'\) is a possibly uncountable index set and for each \(j' \in J'\), \(U_{j', j} \subseteq T_j\) s are some open subsets only some finite of which are not \(T_j\) s, by Note for the definition of product topology.
Step 2:
\(\pi^l (U) = \pi^l (\cup_{j' \in J'} \times_{j \in J} U_{j', j}) = \cup_{j' \in J'} \pi^l (\times_{j \in J} U_{j', j})\), 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, \(= \cup_{j' \in J'} U_{j', l}\), which is open on \(T_l\).
So, \(\pi^l\) is open.
