0
$\begingroup$

As suggested by the title, I have to prove that (for any topological space $X$) if every open subspace of $X$ is normal, then $X$ is completely normal. I proved a similar exercise ($X$ completely normal implies that every subspace of $X$ is completely normal), but I'm having trouble with this one.

Def. A topological space $X$ is completely normal if, whenever $A, B \subseteq X$ with $A\cap \overline{B}=\emptyset$ and $\overline{A}\cap B=\emptyset$, there are disjoint open sets $U,V\subseteq X$ with $A\subseteq U$ and $B\subseteq V$.

Any suggestions would be welcome.

$\endgroup$
  • $\begingroup$ math.stackexchange.com/questions/290279/… $\endgroup$ – user327401 Apr 1 '16 at 16:02
  • $\begingroup$ It looks like their argument does not need to be modified. I would have thought that, since my statement is stronger, something more would be needed. Thanks! $\endgroup$ – Reigh Apr 1 '16 at 16:38
1
$\begingroup$

Suppose $A$ and $B$ are completely separated in $X$, i.e. $\operatorname{cl}_X(A) \cap B = A \cap \operatorname{cl}_X(B) = \emptyset$. If we'd know their closures were disjoint (which we don't!) we could use normality even in $X$ (separating the closures), but we cannot.

So define $Y = X \setminus (\operatorname{cl}_X(A) \cap \operatorname{cl}_X(B))$, cutting out the potential problem set. Then $Y$ is open in $X$, and so $Y$ is normal by assumption. Also $\operatorname{cl}_Y(A)$ and $\operatorname{cl}_Y(B)$ are disjoint in $Y$ (check this !). This can be shown using the general fact that $\operatorname{cl}_Y(Z) = \operatorname{cl}_X(Z) \cap Y$ for all subsets $Z$ of $X$, and $A$ and $B$ being completely separated in $X$. Now apply normality of $Y$ to these closures in $Y$ to finish the proof.

$\endgroup$
  • $\begingroup$ X is completely normal iff X is hereditarily normal. Hereditarily normal means every subspace is normal. $\endgroup$ – DanielWainfleet Apr 2 '16 at 8:56
  • $\begingroup$ @user254665 we cannot assume the statement we are trying to prove here... This is part of the proof of your statement. $\endgroup$ – Henno Brandsma Apr 2 '16 at 9:37
  • $\begingroup$ I'm not suggesting we assume it to solve the Q. I was mentioning it as a further result. $\endgroup$ – DanielWainfleet Apr 2 '16 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.