A description/proof of that map that is anywhere locally constant on connected topological space is globally constant
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that any map that is anywhere locally constant on any connected topological space is globally constant.
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: Description
For any connected topological space, \(T\), and any map, \(f: T \rightarrow S\) where \(S\) is any set, such that around each point, \(p \in T\), there is an open set, \(U_p \subseteq T, p \in U_p\), such that \(f\) is constant on \(U_p\), \(f\) is constant globally.
2: Proof
The set of \(U_p\)s for all the points constitutes an open cover, \(S_c = \{U_p\}\).
Choose any point, \(p_0 \in T\), and there is the open set, \(U_{p_0} \in S_c, p_0 \in U_{p_0}\), on which \(f\) takes the constant image, \(f (p_0)\). For any other point, \(p \in T\), there is the open set, \(U_{p} \in S_c, p \in U_{p}\), on which \(f\) takes the constant image, \(f (p)\).
By the proposition that any pair of elements of any open cover of any connected topological space is finite-open-sets-sequence-connected via some elements of the open cover, there is a sequence of elements of \(S_c\) starting at \(U_{p_0}\) and ending at \(U_p\) where each adjoining pair of elements of the sequence shares a point. Obviously, via those shared points, \(f\) takes the same image, \(f (p_0)\), on all the elements of the sequence. So, \(f (p) = f (p_0)\).
3: Note
Although the proposition may seem obvious, we have to exactly prove the existence of such a sequence.
