description/proof of that connected component is quasi-connected component on locally connected topological space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally connected topological space.
- 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 for any topological space, each connected component is contained in the corresponding quasi-connected component.
- The reader admits the proposition that any quasi-connected component is open on any locally connected topological space.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
- The reader admits the proposition that for any topological space, each quasi-connected component is closed.
- The reader admits the proposition that any constant map between any topological spaces is continuous.
- The reader admits the proposition that any quasi-connected topological space is connected.
Target Context
- The reader will have a description and a proof of the proposition that any connected component is the corresponding quasi-connected component on any locally connected topological space.
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:
\(T\): \(\in \{\text{ the locally connected topological spaces }\}\)
\(C\): \(\in \{\text{ the connected components of } T\}\)
\(C'\): \(\in \{\text{ the quasi-connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
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Statements:
\(C = C'\)
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2: Proof
Whole Strategy: Step 1: see that \(C \subseteq C'\) and \(C'\) is uniquely determined; Step 2: see that \(C'\) is open; Step 3: see that \(C'\) as the topological subspace is quasi-connected and \(C'\) is connected; Step 4: conclude the proposition.
Step 1:
\(C \subseteq C'\), by the proposition that for any topological space, each connected component is contained in the corresponding quasi-connected component.
\(C'\) is uniquely determined, because \(C'\) is an equivalence class and is a part of the division of \(T\).
Step 2:
\(C' \subseteq T\) is open, by the proposition that any quasi-connected component is open on any locally connected topological space.
Step 3:
Let us see that \(C'\) as the topological subspace is quasi-connected.
Note that \(C'\)'s being a quasi-connected component is different from \(C'\)'s being a quasi-connected topological space in general, because for the former, we think of the set of the continuous maps from \(T\) into any discrete topological spaces while for the latter, we think of the set of the continuous maps from \(C'\) into any discrete topological spaces.
Step 3 Strategy: Step 3-1: take the set of the continuous maps from \(C'\) into any discrete topological spaces, \(F\), and for each \(f: C' \to T' \in F\), extend \(f\) to a continuous \(f': T \to T'\); Step 3-2: see that for each \(c'_1, c'_2 \in C'\), \(f (c'_1) = f' (c'_1) = f' (c'_2) = f (c'_2)\).
Step 3-1:
Let us take the set of the continuous maps from \(C'\) into any discrete topological spaces, \(F\).
Let \(f: C' \to T' \in F\) be any.
Let \(t' \in T'\) be any.
Let us take the extension of \(f\), \(f': T \to T', t \mapsto f (t) \text{ when } t \in C'; \mapsto t' \text{ otherwise }\).
\(f'\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous: \(\{C', T \setminus C'\}\) is an open cover of \(T\), by the proposition that for any topological space, each quasi-connected component is closed, and \(f' \vert_{C'}: C' \to T' = f\) is continuous and \(f' \vert_{T \setminus C'}: T \setminus C' \to T'\) is continuous, because it is constant, by the proposition that any constant map between any topological spaces is continuous.
Step 3-2:
For each \(c'_1, c'_2 \in C'\), \(f (c'_1) = f' (c'_1) = f' (c'_2) = f (c'_2)\), because \(C'\) is a quasi-connected component.
That means that \(C'\) is a quasi-connected topological subspace.
Step 4:
So, \(C'\) is a connected topological subspace, by the proposition that any quasi-connected topological space is connected.
So, \(C' \subseteq C\), because each pair of points on \(C'\) is connected, so, all the points of \(C'\) are in the same connected component.
So, \(C = C'\).
