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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.

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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.

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  • $\begingroup$ But why would such $U'\supset U$ and $V'\supset V$ be disjoint ? $\endgroup$ – user254885 May 31 at 15:12
  • $\begingroup$ You did not require that. $\endgroup$ – KP Hart May 31 at 15:27
  • $\begingroup$ Indeed. Sorry. Now the question is stated correctly, thanks. $\endgroup$ – user254885 May 31 at 15:32
  • $\begingroup$ 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. $\endgroup$ – user254885 May 31 at 17:24

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