# About “Each regular paratopological group is completely regular” article

In this blog post I consider my attempt to rewrite the article “Each regular paratopological group is completely regular” by Taras Banakh, Alex Ravsky in a more abstract way using my theory of reloids and funcoids.

The following is a general comment about reloids and funcoids as defined in my book. If you don’t understand them, restrict your mind to the special case ${f}$ to be a quasi-uniform space and ${(\mathsf{FCD}) f}$ is the corresponding quasi-proximity.

${\langle f \rangle^{\ast}}$ is the closure operator corresponding to a funcoid ${f}$. I also denote the image ${\mathscr{P} X \rightarrow \mathscr{P} Y}$ of a function ${f : X \rightarrow Y}$ as ${\langle f \rangle^{\ast}}$.

I will also denote ${f^{\circ}}$ the interior funcoid for a co-complete funcoid ${f}$ (for the special case if ${f}$ is a topological space ${f^{\circ}}$ is the interior operator of this space). It is defined in the file addons.pdf (not yet in my book).

By definition (slightly generalizing the special case if ${f}$ is a quasi-uniform space) an endo-reloid ${f}$ on a set ${U}$ is normal when ${\langle (\mathsf{FCD}) f \rangle^{\ast} A \sqsubseteq \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle F \rangle^{\ast} A}$ for every entourage ${F \in \mathrm{up}\, f}$ of ${f}$ and every set ${A \subseteq U}$.

Then it appear “obvious” that this definition of normality is equivalent to the formula:

$\displaystyle (\mathsf{FCD}) f \sqsubseteq ((\mathsf{FCD}) f)^{\circ} \circ (\mathsf{FCD}) f \circ (\mathsf{FCD}) f.$

However, I have failed to prove it. Here is my attempt

$\langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} A = \\ \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} \langle F \rangle^{\ast} A = \\ \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle F \rangle^{\ast} A = ? ?$

The further step fails because in general ${\langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} S \neq \bigsqcap_{F \in \mathrm{up} f} \langle \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \rangle^{\ast} S}$.

So as now my attempt has failed. Please give me advice how to overcome this shortcoming of my theory.