Let $X$ be a compeletly regular topological space with the property that, if $x_1, x_2, ...,x_n,...$ are distinct and arbitrary points of $X$ and if $f:X \rightarrow \mathbf{R}$ is an arbitrary continuous real function with $f (x_i)=0$ for some $i$, then there exists $j\not= i$ with $f(x_j)=0$. I think this space is finite. Is that true?

Note that:

$(X,\mathcal{T})$ is said to be completely regular if it is Hausdorff and for every $x \in X$ and every closed set $C \subseteq X$ not containing x can be separated by a continuous function, that is, $\exists$ a continuous function $f : X \rightarrow [0, 1]$ such that $f(x) = 0$ and $f(a) = 1 \ \forall a \in C$.

  • $\begingroup$ The finite case avoids the problem by making the hypothesis of the property void (you can't choose an infinite sequence of pairwise distinct points). You however should clarify a bit what you mean by "completely regular topological space": for some authors this implies this space is Hausdorff, and for some this does not. For your question this difference is significant. $\endgroup$ Commented Sep 23, 2016 at 13:46
  • $\begingroup$ Let me double check: Your property says that the zero set of every continuous real valued function on $X$ is either finite or uncountably infinite. Is that correct? $\endgroup$
    – Lee Mosher
    Commented Sep 23, 2016 at 14:18
  • $\begingroup$ Hint: Use Tietze extension theorem to prove your conjecture. $\endgroup$ Commented Sep 23, 2016 at 14:20

1 Answer 1


Suppose $X$ is infinite, and let $x_0, x_1, x_2, \dots$ be any sequence of distinct points. Then for each $n \ge 1$ there is a continuous $f_n : X \to [0,1]$ with $f_n(x_0) = 0$ and $f_n(x_n) = 1$. Now the function $f = \sum_{n=1}^\infty 2^{-n} f_n$ is continuous (by uniform convergence) and has $f(x_0) = 0$ but $f(x_n) > 0$ for all $n \ge 1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .