# example of non-complete lattice homomorphism

The real number line $[-\mathrm{\infty},\mathrm{\infty}]=\mathbb{R}\cup \{-\mathrm{\infty},\mathrm{\infty}\}$ 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 $f:[-\mathrm{\infty},\mathrm{\infty}]\to [-\mathrm{\infty},\mathrm{\infty}]$ as

$$f(x)=\{\begin{array}{cc}\hfill 0\hfill & \hfill x\le 0\hfill \\ \hfill 1\hfill & \hfill x>0.\hfill \end{array}$$ |

First notice that if $x\le y$ then $f(x)\le f(y)$, for either $x\le y\le 0$ in which case $f(x)=0=f(y)$, or $$ which gives $$ or $$ 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 $f(s)\le f(t)$ for all $t\in S$,
so $f(\mathrm{min}S)=f(s)=\mathrm{min}f(S)$. Hence $f$ is a lattice homomorphism^{}.

However, $f$ is not a complete lattice homomorphism. To see this let $$. Then $infS=0$. However, $f(infS)=f(0)=0$ while $inff(S)=inf\{1\}=1$.

Title | example of non-complete lattice homomorphism |
---|---|

Canonical name | ExampleOfNoncompleteLatticeHomomorphism |

Date of creation | 2013-03-22 16:58:36 |

Last modified on | 2013-03-22 16:58:36 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 4 |

Author | Algeboy (12884) |

Entry type | Example |

Classification | msc 06B05 |

Classification | msc 06B99 |

Related topic | ExtendedRealNumbers |