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I'm looking for classes of spaces $X$ having the property that for each $x_0 \in X$ there is a continuous map $f:X \to \mathbb R$ such that $Z(f) := f^{-1}(0) = \lbrace x_0\rbrace$. Examples are:

  • Metric spaces with $f(x) = d(x,x_0)$
  • Discrete spaces

Do you know of other classes of spaces with this property ?

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Note that all discrete spaces are metric, so you really have only one class here. –  Brian M. Scott May 5 '12 at 8:29
    
What about completely regular $T_1$-spaces? –  Martin Sleziak May 5 '12 at 8:30
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@Martin: They don't all have the property: look at $\omega_1+1$ with the order topology, for instance. The property implies that every singleton is a $G_\delta$. –  Brian M. Scott May 5 '12 at 8:32
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Perfectly normal Hausdorff spaces have this property, more or less by definition. –  Chris Eagle May 5 '12 at 9:00
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2 Answers 2

up vote 5 down vote accepted

Every submetrizable space has the desired property, where a space $\langle X,\tau\rangle$ is submetrizable iff there is a coarser metrizable topology $\tau'$ on $X$, i.e., a metrizable topology such that $\tau\supseteq\tau'$. More generally, if $\langle X,\tau\rangle$ is any space with the property in question, and $\tau'\supseteq\tau$ is a finer topology on $X$, then $\langle X,\tau'\rangle$ also has the property. The proof is trivial: every $\tau$-continuous $f:X\to\Bbb R$ is automatically $\tau'$-continuous. This provides an abundance of non-metrizable examples, many of which aren't even regular.

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If the space $X$ is completely regular and a point $x_0\in X$ is a $G_\delta$-point. then this point is a Zero set of $X$.this means that each completely regular space with the additional property that, every point of it is a $G_\delta$ is in the class you asked.

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