Tarski Fixed Point Theorem Counterexample?

Question

I have a quick question regarding Tarski's fixed point theorem. It states that for all order-preserving $f:X\to X$, the set of fixed points must be a complete lattice. I was wondering how that would apply for this function: $$f(x) = \begin{cases}0 & x\in\left[0,\frac12\right] \\ x & x\in\left(\frac12,1\right]\end{cases}$$ so that $f:[0,1]\to[0,1]$. But the set of fixed points would be $\mathcal{F} = \{0\}\cup \left(\frac12,1\right]$. However, $\mathcal{F}$ is not a complete lattice, as the subset $\left(\frac12,1\right]$'s infimum is $\frac12$ but $\frac12\not\in\mathcal{F}$. I'm not sure what I'm misinterpreting in either the statement of the theorem, why my counter does not apply, or if I'm misunderstanding the definition of a complete lattice. For the record, $\mathcal{F}$ does have a max and a min fixed point ($0, 1$ namely), but the statement is related to complete lattices and hence my question. Thanks!

Answer 1

In $\{0\}\cup(\frac12,1]$, $0$ is a lower bound of $(\frac12,1]$. It is also the only (and hence greatest) lower bound of $(\frac12,1]$ in $\{0\}\cup(\frac12,1]$.