Suppose such $f$ exists. $A=f^{-1}(\{0\})$ must be closed, and, using exercise 1, it is the countable intersection of open sets. The other direction. If $A$ is closed and there is a countable collection of open sets $V_n$ such that their intersection is $A$ , then let $U_1=V_1$ and for every $n$ if $U_{1/n}$ is defined let $A\subseteq U_{1/(n+1)}\subseteq\overline{U}_{1/(n+1)}\subseteq U_{1/n}\cap V_{n+1}$ . This is possible because the space is normal and $A$ is closed. We need to slightly modify the construction in the proof of the Urysohn lemma. Namely, we do not define $U_0$ , we define a sequence of point of $\mathbb{Q}\cap[0,1]-\{0\}$ starting from 1 such that no $U_p$ is defined before $U_{1/n}$ if $p<1/n$ . So, first we define $U_1=V_1$ , then $U_{1/2}$ , then some $U_p$ for $1/2<p<1$ , then $U_{1/3}$ , then some $U_p$ for $1/3<p<1$ , etc. This way we can define $U_p$ for all rational points in $(0,1]$ with the properties required by the proof. Moreover, $A\subseteq f^{-1}(0)\subseteq\cap_{p>0}U_p=A$ .