description/proof of that for topological space, connected component is contained in quasi-connected component
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological connected component.
- The reader knows a definition of topological quasi-connected component.
- The reader admits the proposition that the preimage of the whole codomain of any map is the whole domain.
- 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.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- The reader admits the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, each connected component is contained in the corresponding quasi-connected component.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(C\): \(\in \{\text{ the connected components of } T\}\)
\(C'\): \(\in \{\text{ the quasi-connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
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Statements:
\(C \subseteq C'\)
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For any fixed \(c \in C\), there is the determined \(C'\) such that \(c \in C'\), and such \(C'\) does not depend on \(c\), by this proposition.
2: Proof
Whole Strategy: Step 1: take any fixed \(c \in C\) and the determined \(C'\) such that \(c \in C'\); Step 2: see that for each \(c' \in C\), for each \(f \in F\), \(f (c') = f (c)\).
Step 1:
Let us take any fixed \(c \in C\).
There is the determined \(C'\) such that \(c \in C'\), which is indeed determined, because \(C'\) is an equivalence class and any set of equivalence classes is a division of the set.
Step 2:
Let \(f \in F: T \to T'\) be any.
\(T = f^{-1} (T')\), by the proposition that the preimage of the whole codomain of any map is the whole domain, \(= f^{-1} (\{f (c)\} \cup (T' \setminus \{f (c)\})) = f^{-1} (\{f (c)\}) \cup f^{-1} (T' \setminus \{f (c)\})\), 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.
\(f^{-1} (\{f (c)\}) \cap f^{-1} (T' \setminus \{f (c)\}) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
\(f^{-1} (\{f (c)\})\) and \(f^{-1} (T' \setminus \{f (c)\})\) are some open subsets of \(T\), because \(T'\) is discrete and \(f\) is continuous.
\(C = T \cap C = (f^{-1} (\{f (c)\}) \cup f^{-1} (T' \setminus \{f (c)\})) \cap C = (f^{-1} (\{f (c)\}) \cap C) \cup (f^{-1} (T' \setminus \{f (c)\}) \cap C)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(f^{-1} (\{f (c)\}) \cap C\) and \(f^{-1} (T' \setminus \{f (c)\}) \cap C\) are some open subsets of \(C\) as the topological subspace and \((f^{-1} (\{f (c)\}) \cap C) \cap (f^{-1} (T' \setminus \{f (c)\}) \cap C) = \emptyset\), because \(f^{-1} (\{f (c)\}) \cap f^{-1} (T' \setminus \{f (c)\}) = \emptyset\).
So, as \(C\) is a connected topological subspace of \(T\), by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger, \(f^{-1} (T' \setminus \{f (c)\}) \cap C = \emptyset\): \(f^{-1} (\{f (c)\}) \cap C \neq \emptyset\), because \(c \in f^{-1} (\{f (c)\}) \cap C\).
That means that \(f (C) \subseteq \{f (c)\}\), which means that for each \(c' \in C\), \(f (c') = f (c)\).
That means that \(C \subseteq C'\).
