description/proof of that quasi-connected topological space is connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of quasi-connected topological space.
- The reader knows a definition of connected topological space.
- 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 any quasi-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:
\(T\): \(\in \{\text{ the quasi-connected topological spaces }\}\)
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Statements:
\(T \in \{\text{ the connected topological spaces }\}\)
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2: Note
In general, a topological quasi-connected component of \(T\) is not any topological connected component of \(T\) (although it contains some topological connected components), but when the topological quasi-connected component is \(T\), it is the connected component, by this proposition.
3: Proof
Whole Strategy: Step 1: suppose that \(T\) was not connected, and find a contradiction; Step 2: as an aside, see that the same logic does not work for when a topological quasi-connected component is not \(T\).
Step 1:
Let us suppose that \(T\) was not connected.
\(T = U_1 \cup U_2\) where \(U_1, U_2 \subseteq T\) were some nonempty open subsets of \(T\) such that \(U_1 \cap U_2 = \emptyset\).
Let us take the discrete \(T' = \{1, 2\}\) and \(f \in F: T \to T', t \mapsto 1 \text{ when } t \in U_1; \mapsto 2 \text{ when } t \in U_2\), which would be indeed continuous, because \(f^{-1} (\emptyset) = \emptyset\), \(f^{-1} (\{1\}) = U_1\), \(f^{-1} (\{2\}) = U_2\), and \(f^{-1} (T') = T\).
Then, there would be a \(u_1 \in U_1\) and a \(u_2 \in U_2\) such that \(f (u_1) = 1 \neq 2 = f (u_2)\), a contradiction against that \(T\) was a topological quasi-connected component: \(u_1\) and \(u_2\) were not topological quasi-connected.
So, \(T\) is connected.
Step 2:
As an aside, let us see why the same logic does not work for when a topological quasi-connected component is not \(T\).
Let \(C \subset T\) be a topological quasi-connected component.
Supposing \(C = U_1 \cup U_2\), taking the discrete \(T' = \{1, 2\}\) and \(f: C \to T', c \mapsto 1 \text{ when } c \in U_1; \mapsto 2 \text{ when } c \in U_2\), \(f\) would not be any map from \(T\) but from \(C\), while \(f\) needed to be a continuous map from \(T\), so, \(f\) would need to be able to be extended on the domain, \(T\), to be continuous, which cannot be proved.
