# Tarski Fixed Point Theorem Counterexample?

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!

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 $$\{0\}\cup(\frac12,1]$$.
• So a complete lattice in $\mathbb{R}$ need not be subcomplete as a sublattice of $\mathbb{R}$? Thanks! May 3, 2020 at 9:30
• @bof Ok, so set of fixed points of $f(S)$ itself is a lattice with the order inherited from $\mathcal{P}(X)$? Also, briefly ignoring completeness, in $\mathbb{R}^n$, are there subsets of $\mathbb{R}^n$ that are lattices in its own right but not a sublattice of $\mathbb{R}^n$? Thanks so much! May 3, 2020 at 10:45
• @bof Also a follow up. Let $\mathcal{F}$ be the set of fixed points of $f(S)$. Then take $X',X''\in\mathcal{F}$, then their join and meet are no longer the union and intersection of them, but rather the "smallest" subspace that contains them both and the "largest" subspace that both of them contain, respectively. Is that correct? May 3, 2020 at 10:50