(a) Yes. A subspace of a subspace of a completely normal space is a subspace of the space, and, therefore, is normal.(c) Yes. If a set is well-ordered then every element has a successor and $\{(a,x]\}$ is a basis at $x$ where $a<x$ or $a=-\infty$ . Therefore, as in the proof of the Theorem 32.4, for a pair of separated sets we can cover each set with such neighborhoods that do not intersect the other set. Moreover, the neighborhoods belonging to one set do not intersect the neighborhoods belonging to the other set.( For any ordered set such that either each point has a successor or each point has a predecessor the proof above works as well: for the latter case we use basis neighborhoods of the form $[)$ . Also, the next exercise states that every linear continuum is normal: a linear continuum is an example of an ordered set such that it satisfies the least upper bound property the same as a well-ordered set does but every point "in the middle" has no successor or predecessor. Note that even though we know that an arbitrary ordered set is normal, its subspace topology is in general finer than the order topology on the subspace, therefore, we cannot immediately conclude that every ordered set is completely normal. However, it is true as well. )(g) Yes. Similarly to (c), we can prove that $\mathbb{R}_l$ is completely normal. Indeed, the proof of the Theorem 32.4 is extremely similar to the one of Example 2 of the previous section: both use the fact that there is a basis at $x$ with sets of the form $[x,)$ or $(,x]$ and both use only the fact that every point in one closed set has such a neighborhood that does not intersect the other set. Then coverings by such basis sets are disjoint automatically. So, as in (c), we obtain the disjoint open neighborhoods for any pair of sets such that neither one contains limit points of the other one.(b) No. $\mathbb{R}_l$ is completely normal (see (g) above), but $\mathbb{R}_l^2$ is not even normal.(d) Yes. Every subspace is metrizable as well, therefore, normal (Theorem 32.2).(e) No. See Example 2.(f) Yes. Every regular second-countable space is normal (Theorem 32.1). Also every its subspace is also regular and second-countable (see Section 31). Therefore, every regular second-countable space is completely normal.