description/proof of that topological quasi-connected component is closed
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological quasi-connected component.
- The reader knows a definition of closed subset of topological space.
- The reader admits the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
- The reader admits the proposition that the intersection of any possibly uncountable number of closed sets or the union of any finite number of closed sets is closed.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, each quasi-connected component is closed.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(C\): \(\in \{\text{ the quasi-connected components of } T\}\)
//
Statements:
\(C \in \{\text{ the closed subsets of } T\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(c \in C\), and see that \(C = \cap_{f \in F} f^{-1} (f (c))\).
Step 1:
Let \(c \in C\) be any.
\(C = \{t \in T \vert \forall f \in F (f (t) = f (c))\}\).
\(C = \cap_{f \in F} f^{-1} (\{f (c))\}\), because for each \(t \in \{t \in T \vert \forall f \in F (f (t) = f (c))\}\), for each \(f \in F\), \(f (t) = f (c)\), so, \(t \in f^{-1} (f (c))\) for each \(f \in F\), so, \(t \in \cap_{f \in F} f^{-1} (\{f (c))\})\); for each \(t \in \cap_{f \in F} f^{-1} (\{f (c))\})\), \(t \in f^{-1} (\{f (c))\})\) for each \(f \in F\), so, \(f (t) = f (c)\) for each \(f \in F\), so, \(t \in \{t \in T \vert \forall f \in F (f (t) = f (c))\}\).
\(\{f (c)\} \subseteq T'\) is closed, because \(T' \setminus \{f (c)\} \subseteq T'\) is open.
So, \(f^{-1} (\{f (c)) \subseteq T\) is closed, by the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
\(\cap_{f \in F} f^{-1} (\{f (c))\} \subseteq T\) is closed, by the proposition that the intersection of any possibly uncountable number of closed sets or the union of any finite number of closed sets is closed.
So, \(C \subseteq T\) is closed.
