description/proof of that for universal net, composition of net before map into another topological space is universal
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any universal net, the composition of the net before any map into any another topological space is universal.
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:
\(D\): \(\in \{\text{ the directed sets }\}\)
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: D \to T\), \(\in \{\text{ the universal nets }\}\)
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(f'\): \(: T \to T'\)
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Statements:
\(f' \circ f \in \{\text{ the universal nets }\}\)
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2: Proof
Whole Strategy: Step 1: let \(S \subseteq T'\) be any, and see that \(f\) is eventually in \(f'^{-1} (S)\) or in \(T \setminus f'^{-1} (S)\); Step 2: see that \(T \setminus f'^{-1} (S) = f'^{-1} (T' \setminus S)\) and that \(f' \circ f\) is eventually in \(S\) or in \(T' \setminus S\).
Step 1:
Let \(S \subseteq T'\) be any.
\(f\) is eventually in \(f'^{-1} (S)\) or in \(T \setminus f'^{-1} (S)\), because \(f\) is universal.
Step 2:
\(T \setminus f'^{-1} (S) = f'^{-1} (T' \setminus S)\), by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset.
When \(f\) is eventually in \(f'^{-1} (S)\), \(f' \circ f\) is eventually in \(S\), because there is a \(d \in D\) such that for each \(d' \in D\) such that \(d \le d'\), \(f (d') \in f'^{-1} (S)\), which implies that \(f' \circ f (d') \in S\).
When \(f\) is eventually in \(T \setminus f'^{-1} (S) = f'^{-1} (T' \setminus S)\), \(f' \circ f\) is eventually in \(T' \setminus S\), because there is a \(d \in D\) such that for each \(d' \in D\) such that \(d \le d'\), \(f (d') \in T \setminus f'^{-1} (T' \setminus S)\), which implies that \(f' \circ f (d) \in T' \setminus S\).
So, \(f' \circ f\) is eventually in \(S\) or in \(T' \setminus S\).
So, \(f' \circ f\) is universal.
