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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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