Yup, it may sound like an inocent question to many of you; but a very good friend of mine is completely baffled in his research about lattices of inflators on a frame. He asked me very kindly to post this on his behalf and here it is. Any fresh ideas or (counter)examples would definitely help him. Thanks a lot! 



Is the complete lattice of inflators on a frame a frame?
I think the answer is `yes'. Let $F$ be a frame and $I(F)$ its set of inflators ordered by the pointwise ordering. I claim that joins and meets are taken pointwise. It suffices to show that if $\lbrace d\_a\rbrace\_{a\in A}$ is a family of inflators, then so is their pointwise join $d$ and their pointwise meet $m$. 


Let's do $d$ first. If $f\in F$, then $f\leq d\_a(f)\leq d(f)$ for any $a\in A$. Order preserving is also clear: if $f\leq f'$ then $d\_a(f)\leq d\_a(f')\leq d(f')$ for all $a\in A$ and hence $d(f)\leq d(f')$. Thus $d\in I(F)$


For meets, $f\leq d\_a(f)$ for all $a\in A$ and so $f\leq m(f)$. If $f\leq f'$, then $m(f)\leq d\_a(f)\leq d\_a(f')$ for all $a\in A$ and so $m(f)\leq m(f')$.


Now it is clear that $I(F)$ is a frame. If $\lbrace d\_a\rbrace\_{a\in A}\subseteq I(F)$ and $c\in I(F)$, then 



> 
> 
> > 
> > $\left(c\wedge \bigvee\_{a\in A}d\_a\right)(f) = c(f)\wedge \bigvee\_{a\in A}d\_a(f) = \bigvee\_{a\in A}(c(f)\wedge d\_a(f))$
> >  $=\left(\bigvee\_{a\in A}(c\wedge d\_a)\right)(f)$
> > 
> > 
> > 
> 
> 
> 


and so $c\wedge \bigvee\_{a\in A}d\_a=\bigvee\_{a\in A}(c\wedge d\_a)$.


Thus $I(F)$ is a frame.


