Which of the following spaces are completely normal? From Munkre books , page no: $205$
Which of the following spaces are completely normal?
$(a)$ A subspace of a completely normal space
$(b)$ The product of two completely normal spaces.
$(c)$ A well-ordered set in the order topology.
$(d)$ A metrizable space.
$(e)$ A compact Hausdorff space.
$(f)$ A regular space with a countable basis.
$(g)$ The space $\mathbb{R_l}$
Any hints/solution
 A: (a) Of course, by definition: if every subspace of $X$ is normal the same holds for every subspace $A$ of $X$ (as a subspace of $A$ is also one of $X$, hence normal).
(b) No, by the example 2 on page 203 ($\omega_1 \times (\omega_1+1)$ in usual modern notation and $S_\Omega \times \overline{S}_\Omega$ in Munkres' notation) and the fact that all sets in the order topology are completely normal:
(c) Yes, follows from 32.4. In fact all subspaces of ordered spaces are normal.
(d) Yes, metrisable spaces are normal, and metrisability is hereditary.
(e) No, as  witnessed by $[0,1]^J$ with $J$ uncountable, which contains the non-normal subspace $(0,1)^J$. You can also use $\overline{S}_\Omega \times \overline{S}_\Omega$ and re-use example (b).
(f) Such spaces are metrisable (if $T_1$, by the Urysohn metrsiation theorem), so yes. Or use that such spaces are normal (even without $T_1$) and that both properties are hereditary. So the proof is slightly different in both cases.
(g) I presume you mean $\mathbb{R}_l$, the lower limit topology on the reals. 
This is indeed completely normal, as it's homeomorphic to the subspace $(0,1) \times \{0,1\}$ of the ordered square (which is ordered hence completely normal).
