Is Alexandroff Duplicate normal? Let $X$ be any topological space and let $A(X)= X\cup X'$ be its Alexandroff Duplicate. If $X$ is normal, is $A(X)$ also normal?
Thanks for any help.
 A: Yes, if $X$ is a normal space, then $A(X)$ is also normal.
I will consider the underlying set of $A(X)$ to be $X \times \{ 0 , 1 \}$, topologised by declaring the basic open sets to be of the following form:


*

*$(U \times \{ 0 , 1 \}) \setminus ( A \times \{ 1 \})$ where $U \subseteq X$ is open and $A \subseteq X$ is finite;

*$\{ \langle x , 1 \rangle \}$ for $x \in X$.


It follows that any subset of $X \times \{ 1 \}$ open in $A(X)$, and $X \times \{ 0 \}$ is a closed subspace of $A(X)$ which is homeomorphic to $X$. For notational ease, given any $B \subseteq A(X)$ denote


*

*$B_0 = \{ x \in X : \langle x , 0 \rangle \in B \}$;

*$B_1 = \{ x \in X : \langle x , 1 \rangle \in B \}$.


Suppose that $F,E$ are disjoint closed subsets of $A(X)$.  Then $F_0$, $E_0$ are disjoint closed subsets of $X$. By normality of $X$ there are disjoint open $U^\prime,V^\prime \subseteq X$ such that $F_0 \subseteq U^\prime$, $E_0 \subseteq V^\prime$.  Then $U = U^\prime \times \{ 0 , 1 \}$, $V = V^\prime \times \{ 0 ,1 \}$ are disjoint open subsets of $A(X)$. It is now easy to show that $$( U \cup F ) \setminus E = ( U \cup ( F_1 \times \{ 1 \} ) ) \setminus E; \\ ( V \cup E ) \setminus F = ( V \cup ( E_1 \times \{ 1 \} ) ) \setminus F$$ are disjoint open subsets of $A(X)$ including $F,E$, respectively.
