368: Map That Is Anywhere Locally Constant on Connected Topological Space Is Globally Constant

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A description/proof of that map that is anywhere locally constant on connected topological space is globally constant

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of connected topological space.
• The reader admits 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.

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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.

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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.

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Main Body

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1: Description

For any connected topological space, $T$, and any map, $f:T\to S$ 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.

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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, $f(p)=f(p_0)$.

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3: Note

Although the proposition may seem obvious, we have to exactly prove the existence of such a sequence.

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References

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