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A topological space $X$ is totally disconnected if the connected components in $X$ are the one-point sets. Also, a $T_1$ topological space $X$ such that for every closed subset $C$ of $X$ and every point $x \in X\setminus C$, there is a continuous function $f:X\rightarrow[0,1]$ such that $f(x)=0$ and $f(C)={1}$ is called completely Regular.

Now let $X$ be a completely regular totally disconnected topological space and $f$ be a real-valued continuous function over $X$( that is, $f\in C(X)$). Now if $A\subseteq X$ such that $f(A)=\{0, 1\}$, how can we define a real-valued continuous function $g$ over $X$ such that $g(X)=\{0,1\}$ and $f(a)=g(a)$ for all $a\in A$?

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  • $\begingroup$ What is the origin of your question? $\endgroup$ – Paul Frost Oct 3 '18 at 12:44
  • $\begingroup$ This cannot always be done. There are even separable metric counterexamples. $\endgroup$ – Henno Brandsma Oct 3 '18 at 16:38
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Your desired property could be reformulated as “every two subsets of $X$ that are functionally separated are clopenly separated”. This property (together with complete regularity) is called strong zero-dimensionality. In the context of Hausdorff spaces we have the following implications: \begin{align}\text{strongly zero-dimensional} \implies \text{zero-dimensional} \implies \text{totally separated} \implies \text{totally disconnected}.\end{align} The implications cannot be reversed. For example the Cantor's leaky tent with the apex removed is a metrizable totally disconnected space that is not totally separated. On the other hand, if your space $X$ is compact, then all the conditions are equivalent.

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  • $\begingroup$ Thanks for your answer. Where can I find a proof in compact case? $\endgroup$ – Es.Ro Oct 3 '18 at 15:25
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    $\begingroup$ @E.Rostami: going from totally separated to strongly zero-dimensional (or even ultranormal) is analogous to proving that every compact Hausdorff space is normal. The last implication follows from the fact that in a normal compact space, components are the same as quasi-components – you can directly prove that quasi-components are connected. $\endgroup$ – user87690 Oct 3 '18 at 15:44
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    $\begingroup$ We will have a counterexample in a totally disconnected separable metric space of dimension $n > 0$. E.g. Erdös space will do as well. $\endgroup$ – Henno Brandsma Oct 3 '18 at 16:37

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