Separating two sets in normal topological spaces by binary relations

Question

Trying to generalize Urysohn lemma or at least to rewrite its proof in a new way:

Let $\mu$ be a normal ($T_4$) topological space on a set $\mho$.

I will denote $\operatorname{up}\mu$ the set of all binary relations $U$ on $\mho$ such that $Ux$ (where $U$ is considered as a multivalued function) is a neighborhood of $x$ for every $x\in\mho$.

Let it is known that binary relation $U_0=P_0\circ Q_0^{-1}$ for $P_0,Q_0\in\operatorname{up}\mu$ and $(U_0\circ U_0^{-1}) \cap A\times B = \emptyset$ for some disjoint closed sets $A$ and $B$ (that is vaguely saying, $A$ and $B$ are "separated" by $U_0$).

Prove (or disprove) existence of binary relation $U_1=P_1\circ Q_1$ where $P_1,Q_1\in\operatorname{up}\mu$ such that $U_0\supseteq U_1\circ U_1^{-1}$ and $(U_1\circ U_1^{-1}) \cap A\times B = \emptyset$.

Hint: A topological space $\mu$ is normal if and only if for every $P\in\operatorname{up}\mu$ there exists $Q\in\operatorname{up}\mu$ such that $Q\circ Q^{-1}\circ Q\circ Q^{-1} \subseteq P\circ P^{-1}$. (I have a proof of this statement but it uses theory of funcoids, which is not widely known.)

Answer 1

Let's find such an $U_1$ that $\neg \left( A \mathrel{[U_1 \circ U_1^{- 1}]^{\ast}} B \right)$:

We have $\neg \left( A \mathrel{[U_0 \circ U_0^{- 1}]^{\ast}} B \right)$.

By normality (see the hint in the question) $\neg \left( A \mathrel{[U' \circ U'^{- 1} \circ U' \circ U'^{- 1}]^{\ast}} B \right)$ for some $U' \in \operatorname{up} \mu$.

Take $Z = U' \circ U'^{- 1}$.

$\neg \left( A \mathrel{[Z \circ Z^{- 1}]^{\ast}} B \right)$.

$U_0 \supseteq P \circ P^{- 1}$ where $P\in\operatorname{up}\mu$, thus by normality $U_0 \supseteq Q \circ Q^{- 1} \circ Q \circ Q^{- 1}$ where $Q\in\operatorname{up}\mu$. Thus there exists $W \in \operatorname{up} \mu$ such that $U_0 \supseteq W \circ W^{- 1} \circ W \circ W^{- 1}$. Thus take $T = W \circ W^{- 1}$ and we have $U_0 \supseteq T \circ T^{- 1}$.

Take $U_1 = Z \cap T$. Because $\mu$ is a topological space (or more generally a complete funcoid (note that it does not hold for arbitary funcoids)), we have $U_1 \in \operatorname{up} \mu$.

$U_0 \supseteq U_1 \circ U_1^{- 1}$ because $U_1 \subseteq T$; $\neg \left( A \mathrel{[U_1 \circ U_1^{- 1}]^{\ast}} B \right)$ because $U_1 \subseteq Z$.