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This is problem $1$, Section $3$, page $252$ of Dugundji's book.

Let $X$ be a completely regular space, $C \subset X$ compact and $U$ an open set containing $C$. Prove there exists a continuous map $f: X \rightarrow [0,1]$ such that $f(c)=0$ for all $c \in C$ and $f(x)=1$ for all $x \in Y \setminus U$.

So the hint is: For each $c \in C$, let $f_{c}$ be such that $f_{c}(c)=0$ and $f_{c}(X \setminus U)=1$, cover $C$ by finitely many sets $\{f_{c_{i}}^{-1}([0,1/2)): i =1,2,..,n\}$ and consider the product $f_{c_{1}}(x)f_{c_{2}}(x)...f_{c_{n}}(x)$.

My question is: the product won't work right? take $z \in C$ then $z \in f_{c_{i}}^{-1}([0,1/2)$, so that $f_{c_{i}}(z) \in [0,1/2)$, but there's no guarantee that $z$ must be one of the $c_{i}'s$ so that we need somehow to modify the function. Can we take the minimum of $\{f_{c_{1}}(x),...,f_{c_{n}}(x)\}$ and then re-scale the function?

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Yes you can. Alternatively you can start with rescaling $f_{c_i}$ so that $f_{c_i}^{-1}[0,1/2)$ is sent to 0 and $f_{c_i}^{-1}(1)$ is sent to 1. (Take any continuous map from $[0,1]$ to itself that sends $[0,1/2)$ to 0 and 1 to 1 and compose it with $f_{c_i}$) Then product would work. – Soarer Aug 13 '11 at 23:16
up vote 3 down vote accepted

Yes, the product requires some modification. You know that for any $x \in C$, $0 \le f_{c_i}(x) < 1$, and there is at least one $i$ such that $0 \le f_{c_i}(x) < 1/2$. It follows that $0 \le \prod\limits_{i=1}^n f_{c_i}(x) < 1/2$ for every $x \in C$. It’s also clear that $\prod\limits_{i=1}^n f_{c_i}(x) = 1$ for every $x \in X \setminus U$. Can you show that the function $$g(x) = 2\max\{f(x),1/2\}-1$$ has the desired properties, where $f(x) = \prod\limits_{i=1}^n f_{c_i}(x)$? Equivalently, compose $f$ with the function $h:[0,1]\to [0,1]$ defined by $h(x) = 0$ if $0 \le x < 1/2$ and $h(x) = 2x - 1$ if $1/2 \le x \le 1$.

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I liked $g(x)$, thanks! – user10 Aug 14 '11 at 1:21

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