# extending disjoint open subsets of a normal Hausdorff space

Under what assumptions on $$C$$ and $$X$$ is the following true ? I was neither able to find a counterxample or prove this, though it appears that compactness, e.g. assuming $$X$$ is compactly generated, may be of help. Is $$X$$ being metrizable helpful?

Let $$C$$ be a closed subset of a normal Hausdorff space $$X$$. Any two open disjoint subsets $$U$$ and $$V$$ of a closed subset $$C$$ of $$X$$ (i.e. $$U$$ and $$V$$ are open in $$C$$) can be "extended" to disjoint open subsets $$U'\supset U'$$ and $$V'\supset V$$ of $$X$$ such that $$U=U'\cap C$$ and $$V=V'\cap C$$, and $$U'\cap V'=\emptyset$$.

Let $$C$$ be a closed subset of a normal Hausdorff $$X$$. For any two closed subsets $$A'$$ and $$B'$$ of $$X$$, any two open subsets $$U\supset A$$ and $$V\supset B$$ of $$C$$ separating $$A=A'\cap C$$ and $$B=B'\cap C$$, i.e. $$U\cap V=\emptyset$$, there exist open subsets $$U'\supset A'$$ and $$V'\supset B'$$ of $$X$$ "extending" $$U$$ and $$V$$, and separating $$A'$$, and $$B'$$, i.e. $$A'\subset U'$$, and $$B'\subset V'$$, and $$U=U'\cap C$$, and $$V=V'\cap C$$, and $$U'\cap V'=\emptyset$$.

The motivation for the question is to clarify this question Closed embedding into a normal Hausdorff space and left lifting property.

The first is true in any hereditarily normal space: separated sets have disjoint neighbourhoods. It fails in the compact product $$(\omega_1+1)\times(\omega+1)$$ (Tychonoff's plank with corner point). The set $$C=\{(\alpha,\beta): \alpha=\omega_1$$ or $$\beta=\omega\}$$ is closed. The sets $$U=\{(\alpha,\omega):\alpha<\omega_1\}$$ and $$V=\{(\omega_1,n):n<\omega\}$$ are open-in-$$C$$ but have no disjoint extensions.

Similarly, in the second statement $$A$$ and $$B$$ are already closed-in-$$X$$, so the second statement is true for hereditarily normal spaces and false for the same example.

Addendum: the first statement characterizes hereditary normality: if $$A$$ and $$B$$ are separated let $$C=\overline{A\cup B}$$ and $$U=C\setminus\overline{B}$$ and $$V=C\setminus\overline{A}$$. Then $$U$$ and $$V$$ are open in $$C$$, with $$A\subseteq U$$ and $$B\subseteq V$$. Then $$U'$$ and $$V'$$ would be disjoint neighbourhoods of $$A$$ and $$V$$ respectively.

• But why would such $U'\supset U$ and $V'\supset V$ be disjoint ? May 31, 2021 at 15:12
• You did not require that. May 31, 2021 at 15:27
• Indeed. Sorry. Now the question is stated correctly, thanks. May 31, 2021 at 15:32
• Thank you! Yes, this is correct, and probably good for our purposes. Unfortunately, my second statement was misstated ($A'$ and $B'$ were intended to be arbitrary), again, but your answer still applies. May 31, 2021 at 17:24