(a) The intersection of $1/n$ -neighborhoods of a closed set (see exercise 2 of §27) is the set itself: since the set is closed, for a point outside it there is a neighborhood that does not intersect the set. Therefore, a metric space being normal is perfectly normal as well.(b) Let $A,B$ be a pair of separated sets, $f$ and $g$ be two continuous functions that vanish precisely on $\overline{A}$ and $\overline{B}$ , respectively. Let $h=f-g$ . It is continuous, $A\subseteq f^{-1}(\mathbb{R}_-)$ and $B\subseteq f^{-1}(\mathbb{R}_+)$ .(c) We need some completely normal space that is not perfectly normal. This should be a normal space such that there is a closed set which is not $G_\delta$ set. What types of completely normal spaces we know so far: metric, ordered and regular second-countable. Metric spaces are perfectly normal. In fact, I looked up ahead, and have realised that the Urysohn Theorem states that a regular second-countable space is metrizable. So, this leaves us with the option to find an ordered space with a closed set that is not $G_\delta$ set. The obvious example is $\overline{S}_\Omega$ . Its largest element is the one that causes all the problems with the space: it is not separable, it is not Lindelöf, it is not second-countable — all because of the basis at $\omega$ . And, I believe, for the same reason it is not perfectly normal. For each set in any countable collection of neighborhoods of $\omega$ find a basis neighborhood $(a_n,\omega]$ contained in the neighborhood. Then $\{a_n\}$ has an upper bound, and the intersection cannot be $\{\omega\}$ .