construction of a discrete family Question: Let $\mathcal{A}=\{A_\alpha : \alpha \in \Lambda\}$ be a discrete family of closed sets in a normal space $X$. Then can we find a discrete family $\mathcal{W}=\{W_\alpha : \alpha \in \Lambda\}$ of regular closed sets in $X$ such that $A_\alpha \subseteq W_\alpha$, for all $\alpha \in \Lambda$.
Definition: (i) $\mathcal{A}$ is said to be discrete family if for each $x\in X$ there is an open set which cuts at most one member of $\mathcal{A}$.
(ii) $A\subseteq X$ is regular closed if $A=cl(int(A))$. Note that $int(cl(A)) =X\setminus cl(int(A))$ and for any open set $A, A\subseteq int(cl(A))\subseteq cl(int(A))$
Observation: (i)In a normal space two disjoint regular closed set can be separated by two disjoint regular open set.
(ii)In a normal space, for a discrete family $\{A_\alpha\}$ of regular closed sets not necessarily one can find a discrete family of open sets $\{O_\alpha\}$ such that $A_\alpha \subseteq O_\alpha$.
That is, if the above result is true that does not hamper normality and collectionwise normality of a space.  (ii) Result is obvious when $\Lambda$ is finite and I guess it can be done upto $\Lambda$ is countable. I can't find what to do when $\Lambda$ is uncountable, please help.
 A: Assuming that the definition of a discrete family is the usual one, that is, $\mathcal A$ is discrete if and only if for each $x \in X$, there is a neighborhood $U$ of $x$ which intersects at most one element of $\mathcal A$, there are at least consistent examples where such a discrete expansion does not exist.  Assume $2^{\aleph_1} = 2^{\aleph_0}$. (This assumption follows, for example, from Martin's Axiom plus the denial of the Continuum Hypothesis.) Then there is a separable normal space $X$ having an uncountable closed discrete subset.  The singletons in the subset form a discrete uncountable family, but since the space is separable, and, therefore, does not have an uncountable collection of disjoint open subsets, there is not a collection $\mathcal W$ of the sort asked for. 
A: 
Lemma. Let $X$ be a topological space, $U\subseteq X$ be open and $D=cl(U)$. Then $cl(int(D))=D$. In other words closures of open subsets are regular closed.

Proof. "$\subseteq$" Let $x\in cl(int(D))$. Then for any open neighbourhood $V$ of $x$ we have $V\cap int(D)\neq\emptyset$. Since $int(D)\subseteq cl(U)$ then $V\cap cl(U)\neq\emptyset$ and thus $x\in cl(U)$.
"$\supseteq$" Note that $U\subseteq D$ and since $U$ is open then $U\subseteq int(D)$. In particular $D=cl(U)\subseteq cl(int(D))$. $\Box$

Let $\mathcal{A}$ be a discrete family of closed subsets. Let me restate the definition of a discrete family (correct me if I'm wrong): there is a family $\mathcal{U}$ of open subsets such that


*

*for each $A\in\mathcal{A}$ there is $U\in\mathcal{U}$ such that $A\subseteq U$

*$U\cap V=\emptyset$ for $U,V\in\mathcal{U}$, $U\neq V$


Pick $A\in\mathcal{A}$ and the corresponding $U\in\mathcal{U}$ (i.e. $A\subseteq U$). Put $F:=X\backslash U$. Then $A\cap F=\emptyset$ and since both are closed (and $X$ is normal) then we can separate them with open neighbourhoods $U_A$ and $U_F$ respectively. Define $W_A:=cl(U_A)$. 


*

*Obviously $A\subseteq W_A$.

*$\mathcal{W}=\{W_A\}_{A\in\mathcal{A}}$ is discrete. Indeed $W_A\cap F=\emptyset$ ($F$ defined earlier) and since both are closed we can separate them by $U_{W_A}$ and $U'_F$. The family $U_{W_A}$ is what we are looking for.

*Since each $W_A$ is a closure of an open subset then it is regular closed by the lemma.

A: If the family $\mathcal A$ is countable then a positive answer follows from a positive answer to this question. 
If the family $\mathcal A$ is uncountable then an answer, in general, is negative even in ZFC. Indeed, let $X$ be the space constructed below in Example J (see [T, $\S $2, p. 713]), the referenced Junilla’s work is [J].

I recall that a space is $\aleph_1$-collectionwise Hausdorff if for each its discrete family $\mathcal{A}=\{\{x_\alpha\} : \alpha <\aleph_1 \}$ of one-point sets  there exists a  family $\mathcal{V}=\{V_\alpha : \alpha<\aleph_1\}$ of disjoint open sets such that $x_\alpha\in V_\alpha$ for each $\alpha<\aleph_1$. A space satisfies countable chain condition provided each its family of non-empty disjoint open sets is (at most) countable. 
Now let $\mathcal{A}$ be the family violating $\aleph_1$-collectionwise Hausdorffness of the space $X$. Assume that $\mathcal{W}=\{W_\alpha : \alpha<\aleph_1\}$ is a discrete a family of regular closed sets in $X$ such that $x_\alpha\in W_\alpha$ for each $\alpha<\aleph_1$. Then $\mathcal{W’}=\{\operatorname{int} W_\alpha : \alpha<\aleph_1\}$ is an uncountable family of non-empty disjoint open subsets of the space $X$, a contradiction. 
References
[J] H.J.K. Junilla Countability of point ﬁnite families of sets, Canad. J. Math. 31 (1979), 673–679.
[T] Gary Gruenhage Normality versus collectionwise normality, p. 685-732 in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.
