description/proof of that for locally path-connected topological space, path-connected component is connected component
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally path-connected topological space.
- The reader knows a definition of topological connected component.
- The reader knows a definition of topological path-connected component.
- The reader admits the proposition that for any topological space, any path-connected component is contained in the corresponding connected component.
- The reader admits the proposition that any path-connected topological component is open and closed on any locally path-connected topological space.
- The reader admits the proposition that for any topological space and any topological subspace that is not necessarily open on the basespace, any subset of the subspace is open on the subspace if it is open on the basespace.
- 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 locally path-connected topological space, any path-connected component is a 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:
\(T\): \(\in \{\text{ the locally path-connected topological spaces }\}\)
\(C\): \(\in \{\text{ the path-connected components of } T\}\)
\(C'\): \(\in \{\text{ the connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
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Statements:
\(C = C'\)
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2: Note
\(C'\) is uniquely determined, by this proposition, because for any other \(\widetilde{C'}\) taken based on \(\widetilde{c} \in C\), \(C = \widetilde{C'}\) means that \(c \in \widetilde{C'}\), so, \(\widetilde{C'}\) is the connected component such that \(c \in \widetilde{C'}\), which is \(C'\).
3: Proof
Whole Strategy: Step 1: see that \(C \subseteq C'\); Step 2: suppose that \(C \subset C'\), and find a contradiction.
Step 1:
\(C \subseteq C'\), by the proposition that for any topological space, any path-connected component is contained in the corresponding connected component.
Step 2:
Let us suppose that \(C \subset C'\).
For each \(c' \in C'\), there would be the path-connected component, \(C_{c'} \subseteq T\), such that \(c' \in C_{c'}\).
\(C_{c'} \subseteq C'\), by the proposition that for any topological space, any path-connected component is contained in the corresponding connected component.
\(C_{c'}\) would be open on \(T\), by the proposition that any path-connected topological component is open and closed on any locally path-connected topological space.
\(C_{c'}\) would be open on \(C'\), by the proposition that for any topological space and any topological subspace that is not necessarily open on the basespace, any subset of the subspace is open on the subspace if it is open on the basespace.
So, there would be the set of some open subsets of \(C'\), \(S := \{C_{c'} \vert c' \in C'\}\), in which \(C\) would be contained.
As the path-connected components were some equivalence classes, any 2 path-connected components would be the same or disjoint.
So, \(S \setminus \{C\}\) would be nonempty and \(C \cap \cup (S \setminus \{C\}) = \emptyset\).
\(\cup (S \setminus \{C\})\) would be open on \(C'\), because it would be the union of some open subsets.
\(C' = \cup S = C \cup \cup (S \setminus \{C\})\).
That would mean that \(C'\) was not connected, a contradiction: refer to the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
So, \(C = C'\).
