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Let $X$ be a connected metric space and $D$ a set. Let $f: X \to D$ such that for every $x \in X$ there is a neighborhood in which $f$ is constant. Show that $f$ is constant.

I'm trying to get the hang of the problem, but this seems somewhat self explanatory? Perhaps I'm not understanding it correctly. If we have that $\forall x \in X:$ $\exists V_\varepsilon(x)$ such that $f$ is constant then wouldn't it imply that it must be constant all in all? The open sets acts as covers for $X$ and since $X$ is connected they cover whole $X$ right?

Ralph
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  • I don't quite understand your argument. Here's a question you should consider: how do you use the connectedness of $X$? That is, in what way does your argument fail in the case that $X$ is not connected? – Ben Grossmann Apr 29 '21 at 15:49
  • An example of a non-constant function $f$ that satisfies the hypothesis for a non-connected $X$: take $X = [0,1] \cup [2,3] \subset \Bbb R$ under the usual topology over $\Bbb R$. Consider the function $f:X \to \Bbb R$ given by $$ f(x) = \begin{cases} 0 & 0 \leq x \leq 1\ 1 & 2 \leq x \leq 3. \end{cases} $$ – Ben Grossmann Apr 29 '21 at 15:52
  • u are right that for all $x$ there exists an open nbhd of $x$ such that $f$ is constant, but what if for $x_1$ and $x_2$, in the nbhd of $x_1$, $f$ is, for example, $5$ and in the nbhd of $x_2$, $f$ is $-2$? – C Squared Apr 29 '21 at 16:38
  • also not sure where you were going with the “open sets acting as a cover for $X$”. this has nothing to do with $X$ being connected – C Squared Apr 29 '21 at 16:45

2 Answers2

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Hint:

If $D$ has only one element, there is nothing to prove, so assume that $D$ has at least two different elements $a$ and $b$ and suppose for the sake of contradiction that there exist some $x,y\in X$ such that $f(x)=a$ and $f(y)=b$. Define two subsets in $X$:

  • Let $U=\{t\in X:f(t)=a\}$.
  • Let $V=X\setminus U$.

Are they empty? Are they open? Is their union $X$? Are the disjoint? What does this imply? Be careful with the second question: this is the crucial point of the problem.

ajotatxe
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Hint Pick some arbitrary $a \in X$ and set $b:=f(a)$.

First show that $f^{-1}(b)$ is closed.

Next, use the fact that that for every $x \in X$ there is a neighborhood in which $f$ is constant to show that $f^{-1}(b)$ is open.

N. S.
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