MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A_n$..?

share|cite|improve this question
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
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
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.