The question is as following:
Design a finite complete lattice $A$, with at least 15 elements. Provide four non-trivial examples of $f:A -> A$ such that $f$ is:
a) non monotone b) monotone but not continuous c) a function with no fixpoints d) continuous.
I am struggling to understand (and then give an example) of continuous/non continuous functions in a lattice. I think that I can deal with parts a and c myself. I am an absolute beginner to lattices, and neither I am studying maths. I checked at some material on the web but I still don't have an idea how you define a continuous function in a lattice.
I would appreciate any answer. So far, I have been just reading for continuous functions on lattices but even finding good references on it is proving to be a challenge.
If you’re using the usual definitions, (b) is impossible. A map $f:A\to A$ is continuous if $f(\sup D)=\sup f[D]$ whenever $D$ is a directed subset of $A$. $D\subseteq A$ is directed if each finite subset of $D$ has an upper bound in $D$. Note that when $D$ itself is finite, this just means that it has a maximum element: the finite directed subsets of $A$ are the finite subsets that have a maximum element.
Suppose that $A$ is finite and $f:A\to A$ is monotone, and let $D$ be a directed subset of $A$. Then $D$ has a maximum element $d_0$, so $f(d)\le f(d_0)$ for each $d\in D$, and therefore $f(d_0)=\max f[D]$. That is, $$f(\sup D)=f(\max D)=f(d_0)=\max f[D]=\sup f[D]\;,$$ and $f$ is therefore continuous.
To get a monotone map that isn’t continuous, you’ll need an infinite complete lattice. Here’s a fairly simple example that may help you to see what has to happen. Let $L=\Bbb N\cup\{\infty\}$, with the usual order $\le$ on $\Bbb N$ together with $a\le\infty$ for each $a\in L$. This is a complete lattice. Let
$$f:L\to L:a\mapsto\begin{cases} 0,&\text{if }a\in\Bbb N\\ \infty,&\text{if }a=\infty\;; \end{cases}$$
clearly $f$ is monotone. Next, note that $\Bbb N$ is a directed subset of $L$: if $F$ is any finite subset of $\Bbb N$, then $\max F$ is an upper bound for $F$ in $\Bbb N$. But
$$f(\sup\Bbb N)=f(\infty)=\infty\ne 0=\sup f[\Bbb N]\;,$$
so $f$ is not continuous.