description/proof of that for map from set into topological space, set of preimages of open subsets is topology
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topology.
- The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any set into any topological space, the set of the preimages of the open subsets is a 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(S\): \(\in \{\text{ the sets }\}\)
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: S \to T\)
\(O'\): \(\in \{\text{ the topologies for } T\}\)
\(O\): \(= \{f^{-1} (U) \vert U \in O'\}\)
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Statements:
\(O \in \{\text{ the topologies for } S\}\)
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2: Proof
Whole Strategy: Step 1: see that \(O\) satisfies the conditions to be a topology.
Step 1:
Let us see that \(O\) satisfies the conditions to be a topology.
1) \(\emptyset \in O\) and \(S \in O\): \(\emptyset \in O'\) and \(\emptyset = f^{-1} (\emptyset) \in O\); \(T \in O'\) and \(S = f^{-1} (T) \in O\).
2) for any \(U_1 \in O\) and any \(U_2 \in O\), \(U_1 \cap U_2 \in O\): \(U_1 = f^{-1} (U'_1)\) and \(U_2 = f^{-1} (U'_2)\) for some \(U'_1, U'_2 \in O'\), \(U_1 \cap U_2 = f^{-1} (U'_1) \cap f^{-1} (U'_2) = f^{-1} (U'_1 \cap U'_2)\), by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets, but as \(U'_1 \cap U'_2 \in O'\), \(U_1 \cap U_2 \in O\).
3) for any \(U_j \in O\) where \(j \in J\) where \(J\) is any index set not necessarily countable, \((\cup_{j \in J} U_j) \in O\): \(U_j = f^{-1} (U'_j)\) for a \(U'_j \in O'\), \(\cup_{j \in J} U_j = \cup_{j \in J} f^{-1} (U'_j) = f^{-1} (\cup_{j \in J} (U'_j))\), 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, but \(\cup_{j \in J} (U'_j) \in O'\), so, \(\cup_{j \in J} U_j \in O\).
