In which topological spaces is every singleton set a zero set?

Solution 1:

Note that if $f^{-1}(0)=\{x\}$ then there exists a sequence of neighbourhoods $U_n $ of $x$ so that $\{x\}=\bigcap_n U_n$. In the particular case that $X$ is compact Hausdorff this implies that $X$ is first countable. So any non-first countable compact topological space does not have that property. Edit:

We can have a similar characterization as 2. If $X$ is a Tychonoff space then $x$ is a zero set if and only if is a $G_\delta$

proof: Suppose $\{x\}=\bigcap_n U_n$ where $U_n$ is a decreasing sequence of open neighbourhoods of $x$. For each $n$ there is a function $f_n: X\to [0,1]$ so that $f_n(x)=0$ and $f_n(y)=1 $ for all $y\in X\setminus U_n$ ($f_n$ exists since $X$ is Tychonoff). Now, the function $g=\sum_n \frac{1}{2^n} f_n$ satisfies that $g^{-1}(0)=\{x\}$. The other implication is trivial.

Solution 2:

This is a partial answer; I will provide an example of a compact Hausdorff space with a singleton that isn't a zero set.

Consider an uncoutable discrete space $X$ and let $X^+$ be its one-point compactification. Clearly $X^+$ is compact Hausdorff.

We claim that the singleton $\{\infty\}$ isn't $G_\delta$ and therefore can't be a zero set. If it were a countable intersection of open sets $U_n$, each of these must have a finite complement. But since we took $X$ to be uncountable, we cannot hope to get $X\setminus\{\infty\}$ as the countable union of finite sets.