# example of non-complete lattice homomorphism

The real number line $[-\infty,\infty]=\mathbb{R}\cup\{-\infty,\infty\}$ is complete in its usual ordering of numbers. Furthermore, the meet of a subset $S$ of $\mathbb{R}$ is the infimum of the set $S$.

Now define the map $f:[-\infty,\infty]\to[-\infty,\infty]$ as

 $f(x)=\left\{\begin{array}[]{cc}0&x\leq 0\\ 1&x>0.\end{array}\right.$

First notice that if $x\leq y$ then $f(x)\leq f(y)$, for either $x\leq y\leq 0$ in which case $f(x)=0=f(y)$, or $x\leq 0 which gives $f(x)=0<1=f(y)$ or $0 so $f(x)=1=f(y)$.

In the second place, if $S$ is a finite subset of $\mathbb{R}$ then $S$ contains a minimum element $s\in S$. So $f(s)\in f(S)$ and $f(s)\leq f(t)$ for all $t\in S$, so $f(\min S)=f(s)=\min f(S)$. Hence $f$ is a lattice homomorphism.

However, $f$ is not a complete lattice homomorphism. To see this let $S=\{x\in\mathbb{R}:0. Then $\inf S=0$. However, $f(\inf S)=f(0)=0$ while $\inf f(S)=\inf\{1\}=1$.

Title example of non-complete lattice homomorphism ExampleOfNoncompleteLatticeHomomorphism 2013-03-22 16:58:36 2013-03-22 16:58:36 Algeboy (12884) Algeboy (12884) 4 Algeboy (12884) Example msc 06B05 msc 06B99 ExtendedRealNumbers