2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

The answer is yes, and a key ingredient is the following theorem due to George Grätzer:

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.)

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, products of affine complete lattices are affine complete.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy