I'm trying to solve the next problem: A topological space $(X,\tau)$ is called completely normal if, and only if, every subspace is normal. Prove that the following conditions are equivalent:
a) $X$ is completely normal.
b) For every $A,B\subseteq X$ with $(\bar{A}\cap{B})\cup(\bar{B}\cap{A})=\emptyset$, there exist disjoint open sets $U,V\in \tau$ such that $A\subseteq U$ and $B\subseteq V$.
Decide 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 with the order topology.
d) A metrizable space.
I've proved the next things: First for the equivalences. Suppose that $X$ is completely normal and let $A,B\subseteq X$ such that $(\bar{A}\cap{B})\cup(\bar{B}\cap{A})=\emptyset$.Then, is clear that $A\subseteq X\setminus\bar{B}$ and $B\subseteq X\setminus\bar{A}$.
Consider $Z=X\setminus\bar{A}\cup X\setminus\bar{B}$. Note that $Z$
is an open set of $X$ ($Z\in\tau$) and, by the definiton of subspace
topology, $\mathrm{Cl}_{Z}A=\bar{A}\cap Z$ and $\mathrm{Cl}_{Z}B=\bar{B}\cap Z$
are closed sets of $Z$ and they are disjoint. Since $X$ is completely
normal, $Z$ is normal and therefore, there exist disjoint open subsets
of $Z$, $U$ and $V$ such that $\mathrm{Cl}_{Z}A\subseteq U$ and
$\mathrm{Cl}_{Z}B\subseteq V$. Since $Z\in\tau,$we have that $U,V\in\tau$.
Moreover, $A\subseteq\bar{A}\cap Z=\mathrm{Cl}_{Z}A$ and $B\subseteq\bar{B}\cap Z=\mathrm{Cl}_{Z}B$.
Hence, $U$ and $V$ are disjoint open sets of $X$ such that $A\subseteq U$
and $B\subseteq V$. Conversely, show that $X$ is completely normal.
Indeed, let $Y$ be a subspace of $X$ and $A,B$ disjoint closed
subsets of $Y$. Then, $A=\bar{A}\cap Y$, $B=\bar{B}\cap Y$ and
$\bar{A}\cap\bar{B}\cap Y=\emptyset$. Note that from this fact, we
have that $\bar{A}\cap B=\emptyset$ and $\bar{B}\cap A=\emptyset$.
Then, by our asumption, there exist $U_{1},V_{1}\in\tau$ such that
$A\subseteq U_{1}$ and $B\subseteq V_{1}$. Taking $U=U_{1}\cap Y$
and $V=V_{1}\cap Y$,we have that $U$ and $V$ are disjoint open
subsets of $Y$ with $A\subseteq U$ and $B\subseteq V$. Hence $Y$
is normal and $X$ is a completely normal space.
Now for the part of deciding which spaces are completely normal I have this:
For part a) the answer is yes, because if $Y$is a subspace of the
completely normal space $X$, then for all subspace $A$ of $Y$,
$A$ is a subspace of $X$ and therefore $A$ is normal.
For part b) I think that the answer is no. I'm considering the Sorgenfry's line $\mathbb{R}_{l}$. In class we proved that $\mathbb{R}_{l}\times\mathbb{R}_{l}$ is not normal. So I tried to show that $\mathbb{R}_{l}$ is a completely normal space but I couldn't. In class we also proved that every regular and Lindeloff space is a normal spcae so I have to prove that every subspace of the Sorgenfrey's line is a Lindeloff space and with this I can conclude, since $\mathbb{R}_{l}$ is regular and every subspace of a regular space is regular. In some pdfs I saw that is true that every subspace of $\mathbb{R}_{l}$ is Lindeloff, but using arguments that we haven't proved. Could you give me some suggestion to prove this fact or may be another example for part b)?
For part c) I have a question. First, I proved that every well ordered set with the order topology is a normal spcae. But can I conclude with this that every subspace is also normal? I ask this because I think that if $Y$is a subspace of the well ordered set $X$, then the subspace topology of $Y$is different to the order topology over$Y$. Could you help me with this?
And finally, for part d), clearly the answer is yes since every metric space is normal and every subspace of a metric space is also a metric space.
Could you please give me some suggestions for the proofs and the questions?
Thanks.
Best Answer
(c) You are correct in thinking that the subspace topology on a subset of a linearly ordered space need not be the order topology on that subset. For instance, if $X$ is an uncountable well-ordered set with the order topology, and $Y$ is the set of points of $X$ that have immediate predecessors in the order, then the subspace topology on $Y$ is discrete, but the order topology is not. However, every linearly ordered space with the order topology is hereditarily normal (or in your terminology completely normal); I gave a proof here; it’s a bit involved, but the basic idea is fairly simple.
(b) The Tikhonov plank is one of the simplest counterexamples. It’s the product of two well-ordered spaces, but it has a subspace (obtained by deleting just one point) that is not normal; you might try working with it. Alternatively, one way to prove that the Sorgenfrey line is hereditarily normal is to show that it can be embedded as a subspace of a linearly ordered space with the order topology. Specifically, let $X=\Bbb R\times\{0,1\}$ with the order topology derived from the lexicographic order on $X$; then the subspace $\Bbb R\times\{1\}$ is homeomorphic to the Sorgenfrey line.