L is a complete lattice, so every subset has a supremum and infimum. In addition, there exists a function $f:L \rightarrow L$ such that $a \leq b$ implies $f(a) \leq f(b)$. Prove that there exists some $k$ in L such that $f(k) = k $
I started by taking the set $S = \{a | f(a) \geq a\}$, but I am now stuck on this.
This is called the Knaster-Tarski Fixpoint Theorem.
Define $A:=\{x\in L\colon x\leq f(x)\}$.
Since $L$ is a complete lattice you can define $\alpha :=\bigvee _LA$.
By definition $\alpha \in L$ so it makes sense to consider $f(\alpha)$.
Note that $A\neq \varnothing$, (why?).
Let's prove that $f(\alpha)$ is an upperbound of $A$. Let $x\in A$. Since $f$ is order-preserving (or monotone), then $f(x)\leq f(\alpha)$. On the other hand, by definition of $A$ it also holds that $x\leq f(x)$, therefore $x\leq f(\alpha)$.
Since $x$ was arbitrary it was proven that $f(\alpha)$ in an upperbound of $A$ and therefore $\color{blue}{\alpha \leq f(\alpha)}$.
Now using this last inequality and the fact that $f$ is order-preserving yields $f(\alpha)\leq f(f(\alpha))$, therefore $f(\alpha)\in A$, thus $\color{blue}{f(\alpha)\leq\alpha}$.
Hence $f(\alpha)=\alpha$.
You should emulate the proof by using meet instead of join. Consider the set $B=\{x\in L\colon f(x)\leq x\}$.