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Suppose X is a normal topological space. Suppose some metric space for example. If {$A_n$}$_{n=1}^{\infty}$ is a collection of pairwise disjoint closed subsets of X, can we find a continuous function on X such that it takes the constant value $n$ on $A_n$..?

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You will need more than just the sets being pairwise disjoint. For instance, if $X$ is the one-point compactification on $\mathbb{N}$, $A_{0}=\\{\infty\\}$ and $A_{n}=\{n\}$ for all $n$, then there is no continuous real-valued function $f$ on $X$ with $f=n$ on $A_{n}$ for all $n$. –  Joseph Van Name Mar 17 '13 at 19:28
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No, take $A_n=\{1/n\}\subset \mathbb{R}$. –  Vahid Shirbisheh Mar 17 '13 at 19:29
    
Hmm, seems you guys beat me to the punch. But why not just make your comments actual answers? –  David White Mar 17 '13 at 19:39
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I wanted to see if Janson A.J. would have edited the question to make it look more like a research question. For instance, he could have replaced "pairwise disjoint" with something like "locally discrete". –  Joseph Van Name Mar 17 '13 at 19:48

2 Answers 2

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As evidenced in the comments and David´s answer, the problem is that even if each $A_n$ is closed, there can be too much "interaction" between the $A_n$'s. A more extreme example would be to take $A_n=\{ q_n \}$ where $\{q_n\}_{n=1}^\infty$ is some enumeration of the rational numbers.

However if $\{A_n \}_{n=1}^\infty$ is a discrete family of closed subsets of the normal space $X$ and $\{c_n \}_{n=1}^\infty$ is any sequence of real numbers, then there is a continuous $f:X \to \mathbb{R}$ such that $f$ takes the constant value $c_n$ in each $A_n$. For this just note that the union of the $A_n$´s is closed in $X$ and each $A_n$ is clopen in this union.

Note: a family $\{A_n \}_{n=1}^\infty$ of subsets of the space $X$ is a discrete family if any $x \in X$ has a neighborhood that intersects at most one of the $A_n$'s.

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Right. Taking singletons of rationals is the best counterexample. Thanks.. –  Janson A.J Mar 19 '13 at 12:13

No. Continuous functions commute with limits of points in a metric space, so just take any convergent sequence of points, i.e. $A_n$ is the $n$-th point in the sequence. Then if your $f$ existed it would have to take value $\infty$ on the limit of the sequence, but that can't happen because $f$ is real valued.

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Actually that was a mistake that I said it takes constant value n on the nth set. Let's choose some bounded values, like 1/n.. –  Janson A.J Mar 17 '13 at 20:12
    
It still won't work even if you have constant value 1/n on $A_{n}$. –  Joseph Van Name Mar 17 '13 at 20:34
    
Another example is if you assign a sequence of values to the $A_n$ which doesn't converge (rather than converging to $\infty$), e.g. $1,-1,1,-1,...$. I don't think there's any hope for the kind of generalization you're asking for. –  David White Mar 17 '13 at 21:40

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