Order preserving functions that do not preserve binary operations

Question

According to Tarski's Fixpoint Theorem for lattices, if I have a complete lattice, $L$, and an order-preserving function, $f:L \to L$, then the set of all fixpoints of $L$ is also a complete lattice.

Now in my case I have a complete lattice that is just a powerset under inclusion. So $\vee := \cup$ and $\wedge:= \cap$. I also have an order-preserving function, so it seems to me that the set of all fixpoints also forms a complete lattice. The issue is that the set of fixpoints is not closed under $\cup$ and $\cap$, so I am struggling to see how the set of fixpoints is a lattice with the same $\vee$ and $\wedge$ as before, as they are no longer binary operations. Don't $\wedge$ and $\vee$ have to return an element in the set in order to be a lattice?

So through Tarski's theorem, are the join and meet on the set of fixpoints not necessarily the same as the join and meet on the original lattice? This seems strange since the function preserves order and the order defines the join and meet I thought. If an order-preserving function does preserve the join and meet, then why does Tarski's theorem "fail" here?

Answer 1

So through Tarski's theorem, are the join and meet on the set of fixpoints not necessarily the same as the join and meet on the original lattice?

Yes, this is exactly what's going on. Let $P$ denote a poset and $f$ denote an endomorphism of $P$. Then:

1. The fixed points $\mathrm{Fix}(f)$ form a subset of $P$.
2. Hence $\mathrm{Fix}(f)$ forms a subposet in a canonical way.
3. If the poset $P$ happens to be a complete lattice, then the poset $\mathrm{Fix}(f)$ will also happen to be a complete lattice, by Knaster-Tarski. But it needn't be a sublattice of the original lattice.

This occurs because order-embeddings between posets needn't preserve meets or joins. Also interesting to note that just because an order-embedding preserves one of meets/joins, does not mean it preserves the other!

This seems strange since the function preserves order and the order defines the join and meet I thought.

Yep; definability does not imply preserved under homomorphism.