description/proof of that path-connected topological space is connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of path-connected topological space.
- The reader knows a definition of connected topological space.
- 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 the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals.
Target Context
- The reader will have a description and a proof of the proposition that any path-connected topological space is connected.
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:
\(T\): \(\in \{\text{ the path-connected topological spaces }\}\)
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Statements:
\(T \in \{\text{ the connected topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: suppose that \(T\) was not connected, and find a contradiction.
Step 1:
Let us suppose that \(T\) was not connected.
There would be some nonempty open subsets, \(U_1, U_2 \subseteq T\), such that \(T = U_1 \cup U_2\) and \(U_1 \cap U_2 = \emptyset\).
Let \(t_1 \in U_1\) and \(t_2 \in U_2\) be any, which would be possible, because \(U_1\) and \(U_2\) were nonempty.
There would be a path, \(\gamma: [0, 1] \to T\), such that \(\gamma (0) = t_1\) and \(\gamma (1) = t_2\).
\(\gamma^{-1} (T) = [0, 1]\), by the proposition that the preimage of the whole codomain of any map is the whole domain.
But \(\gamma^{-1} (T) = \gamma^{-1} (U_1 \cup U_2) = \gamma^{-1} (U_1) \cup \gamma^{-1} (U_2)\), 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, where \(\gamma^{-1} (U_1) \cap \gamma^{-1} (U_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint, \(\gamma^{-1} (U_1), \gamma^{-1} (U_2) \subseteq [0, 1]\) were open, because \(\gamma\) was continuous, and \(\gamma^{-1} (U_1)\) and \(\gamma^{-1} (U_2)\) were nonempty, because \(0 \in \gamma^{-1} (U_1)\) and \(1 \in \gamma^{-1} (U_2)\), which would mean that \([0, 1]\) was not connected.
But \([0, 1]\) was a connected topological space, by the proposition that the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals, a contradiction.
So, \(T\) is connected.
