An exercise from John Lee's Introduction to Topological Manifolds:

Suppose $X$ is a topological space, and for every $p\in X$ there exists a continuous function $f:X \to \mathbb{R}$ such that $f^{-1}(0)=\{p\}.$ Show that $X$ is Hausdorff.

I'm struggling to get started on this one.

  • 3
    $\begingroup$ Hausdorff means you can separate points by neighbourhoods. Can you use this function $f$ to separate the point $\{p\}$ from some arbitrary point? $\endgroup$ – Ian Coley Feb 16 '14 at 5:53

Hint Consider $q\neq p$. $f(q)\neq 0$, separate $0$ and $f(q)$ and take preimages.


I believe the solution is to take $f(q)=\epsilon$, and then $f^{-1}(-\epsilon/3, \epsilon/3)$ and $f^{-1}(2\epsilon/3, 4\epsilon/3)$ will separate the two points $p$ and $q$.

  • $\begingroup$ Are you answering your own question? Anyway, this is kind of what I and Ian both suggested. $\endgroup$ – Vadim Feb 16 '14 at 6:18


Consider distinct points $p_1, p_2 \in X$. We want to show there exists neighborhoods $U_1 \subset X$ and $U_2 \subset X$ for $p_1$ and $p_2$, respectively, such that $U_1 \cap U_2 = \varnothing$.

Consider the continuous function $f$ such that $f^{-1}(\{0\})=\{p_1\}$, which exists by hypothesis. Then $f(p_1)=0$ and $f(p_2) =r\ne 0$. Since $\mathbb R$ is Hausdorff, there exists disjoint open neighborhoods $V_1$ for $0$ and $V_2$ for $r$ such that $V_1 \cap V_2 =\varnothing$. Now, $p_1 \in f^{-1}(V_1)$, $\ p_2 \in f^{-1}(V_2)$ and since $f$ is continuous $f^{-1}(V_1)$ and $f^{-1}(V_2)$ are open in $X$. Furthermore,

$$f^{-1}(V_1)\cap f^{-1}(V_2) = f^{-1}(V_1\cap V_2)=f^{-1}(\varnothing)=\varnothing.$$

So, $f^{-1}(V_1)$ and $f^{-1}(V_2)$ are disjoint open neighborhoods of $p_1$ and $p_2$. Thus, $X$ is Hausdorff.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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