A $k$-ary function $f$ on a bounded distributive lattice
$L$ is called compatible if for any congruence relation $\theta$ on $L$ and $(a\_i, b\_i)\in \theta$ for $i=1,\ldots,k$ we always have $(f(a\_1,\ldots,a\_k), f(b\_1,\ldots,b\_k))\in\theta$.


It's easy to see that all polynomials are compatible.


A lattice $L$ is called *affine complete* if conversely every compatible function on
$L$ is a polynomial.


Given a cardinal $\kappa > 0$, is $[0,1]^\kappa$ an affine complete lattice?



Is $[0,1]^\kappa$ an affine complete lattice?
The answer is yes, and a key ingredient is the following theorem due to George Grätzer:



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> > A bounded distributive lattice is affine complete if and only i
> >  f it does not
> >  contain a proper interval that is a Boolean lattice in the ind
> >  uced order. (G. Grätzer, *Boolean functions on distributive lattices*, Universal Algebra and Applications, vol.9, Banach Center Publications, Warsaw, 97-104.)
> > 
> > 
> > 
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> 


First we show that $[0,1]$ is affine-complete: Given any interval $[a,b]\subseteq [0,1]$ with $a<b$, it's easy to see that $\frac{a+b}{2}$ doesn't have a completement, therefore $[0,1]$ doesn't contain any proper Boolean intervals.


Second, according to Theorem 4.1. of [this note](http://arxiv.org/pdf/0705.4286.pdf), products of affine complete lattices are affine complete.


