Let $(X,\mathscr {T})$ be a normal (not necessarily hausdorff) space and $Y \subseteq X$ be closed. Let $(Y,\mathscr{T}_Y)$ be $Y$ equipped with the subspace topology of $X$.
If $A$ and $B$ are disjoint and closed in $X$ and there exist disjoint $U',V' \in \mathscr{T}_Y$, such that:
\begin{align*}
A\cap Y\subset U'\\
B\cap Y\subseteq V'\\
\end{align*}
Then we have disjoint $U,V \in \mathscr{T}$, such that:
\begin{align*}
A&\subseteq U\\
B &\subseteq V\\
U'&=U\cap Y\\
\end{align*}
How would I go about proving this? (Also I am thinking the author might have missed $V'=V\cap Y$ in the to-prove-section of the exercise, it seems to fit)
Since $Y \in \mathscr{T}^c$ we know that $A\cap Y$ and $B\cap Y$ are closed in $Y$ iff they are closed in $X$, which is the case, since they are the intersection of two closed sets in $X$. Everything else I've tried doesn't seem useful. Any tricks I'm missing?
Best Answer
Define $U'' = U' \cup (X \setminus Y)$. Then obviously $A \subset U''$ and $Y \cap U'' = U'$. Moreover, $U'' \in \mathscr T$. To see this, choose $U''' \in \mathscr T$ such that $Y \cap U''' = U'$. Then $U'' = U''' \cup (X \setminus Y)$ and the union on the right hand side if open in $X$.
Since $A' = Y \setminus V'$ is closed in $Y$ and $Y$ is closed in $X$, we see that $A'$ is closed in $X$. Thus also $A'' = A \cup A'$ is closed in $X$. By construction $A'' \cap B = \emptyset$.
Now choose $U^*, V \in \mathscr T$ such that $A'' \subset U^*, B \subset V, U^* \cap V = \emptyset$ and define $U = U'' \cap U^* \in \mathscr T$. We have
(1) $A \subset U$
(2) $U \cap V = \emptyset$
(3) $U' \subset A' \subset A'' \subset U^*$, therefore $Y \cap U = Y \cap U'' \cap U^* = U' \cap U^* = U'$.
Note that in general it cannot be expected that $Y \cap V = V'$.