Let $X$ be a topological space and $\{C_i\}_{i\in I}$ be the family of its connected components. Now let $A\subseteq I$, define $f:X\rightarrow \mathbb{R}$ by $f(\cup_{i\in A}C_i)=1$ and $f(X\setminus\cup_{i\in A}C_i)=0$. How can we prove that $f$ is continuous
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$\begingroup$ that's just writing down the definition. Have you tried anything? $\endgroup$– David JaramilloAug 24, 2018 at 1:43
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$\begingroup$ Maybe try it for just two components first $\endgroup$– mheldmanAug 24, 2018 at 1:50
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1 Answer
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Actually, it’s not true.
Take $\Bbb Z_p$, the $p$-adic integers, which is totally disconnected, so that the only connected components are the singleton sets $\{p\}$. In this case, your $I$ is in natural correspondence with the space $\Bbb Z_p$ itself. Now let $A=I\setminus\{0\}$, whose union is, of course, the set of nonzero elements of $\Bbb Z_p$. Defining $\,f$ as you have done, we get a function that’s $1$ everywhere but at zero, where $f(0)=0$. Discontinuous, since zero is in the closure of the other set.
(Lots of other totally disconnected spaces would work as well. Like $\Bbb Q$, for instance.)
