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!
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4$\begingroup$ It might help to define inflators. $\endgroup$– Benjamin SteinbergCommented Oct 11, 2012 at 1:45
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2$\begingroup$ Sure, an inflator in a frame $F$ is a function $d:F\rightarrow F$ such that $d$ is order preserving and inflatory, this meaning that $a\leq d(a)$ holds for all $a\in F$. $\endgroup$– Le FrankCommented Oct 11, 2012 at 3:08
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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.
answered Oct 11, 2012 at 14:44
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$\begingroup$ OK, let me point out something else. This friend of mine just checked up what you did and said something like "oh yes, I had already thought THAT, but I have a feeling that there is something wrong lurking in there." Since he's not comfortable with all this at 100% (and I'm getting tired of writing on his behalf!) Do you have any references where he can strive to get confident about what you just did? Yours and yours alone are this well earned 15 points! Cheers. $\endgroup$– Le FrankCommented Oct 12, 2012 at 1:21
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$\begingroup$ I don't have a reference. I just worked it out. But it seems straightforward enough. $\endgroup$ Commented Oct 12, 2012 at 13:29