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

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.

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 subsets $U$ and $V$ of a closed subset $C$ of $X$ (i.e. $U$ and $V$ are open in $C$) can be "extended" to open subsets $U'\supset U'$ and $V'\supset V$ of $X$ such that $U=U'\cap C$ and $V=V'\cap C$.

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

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

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 subsets $U$ and $V$ of a closed subset $C$ of $X$ (i.e. $U$ and $V$ are open in $C$) can be "extended" to open subsets $U'\supset U'$ and $V'\supset V$ of $X$ such that $U=U'\cap X$ and $V=V'\cap C$.

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

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

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 subsets $U$ and $V$ of a closed subset $C$ of $X$ (i.e. $U$ and $V$ are open in $C$) can be "extended" to open subsets $U'\supset U'$ and $V'\supset V$ of $X$ such that $U=U'\cap C$ and $V=V'\cap C$.

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