Definition of continuous functions in order theory

Question

If we have a complete partial order (i.e. directed complete) I find frequently the following definition of a continuous function. A function $f:A\to B$ where $A$ and $B$ are cpos is called continuous if it maps the suprema of directed subsets of $A$ (if exist) to the corresponding suprema of directed subsets of $B$.

In complete lattices I would define continuous functions as functions which preserve suprema and infima (since both exist in a complete lattice for any subset).

Since complete lattices are cpos the following question arises: Are both definitions consistent? The requirement that all suprema and infima are preserved is stronger than the requirement that only suprema of directed sets are preserved. Therefore it might be possible that both definitions are different. Or are they equivalent?

Answer 1

To make some sense of the various possibilities for maps in order theory, it is best to look at structures in the sense of algebra, rather than properties of maps.

According to algebra, when dealing with structured sets, the corresponding notion of homomorphism should be a map which preserves structure. And then category theory teaches us that the morphisms are just as important as the object (in fact, they are more important).

For example, it may seem odd to distinguish between posets which have all suprema and posets which have all suprema and infima. After all, any poset which has all suprema has the infima too. But the difference matters when we look at the two categories:

1. $\mathbf{SupLat}$: the objects are posets with arbitrary suprema (which are just complete lattices), the morphisms are maps which preserve suprema.
2. $\mathbf{SupInfLat}$: the objects are posets with arbitrary suprema and inifma (which again are just complete lattices), the morphisms are maps which presrerve suprema and infima.

As was already pointed out in Zhen's answer, there are maps on complete lattices which preserve suprema but not infima, so the distinction is meaningful.

You ask about a clear definition of continuity in complete lattices. Continuity is about topology, so we should look at ways of topologizing complete lattices, or more generally posets, as then it will be clear what the continuous maps are. Among all the possibilities, it is probably desirable to restrict to those that allow us to recover the partial order from the topology by passing to the specialization order. The strongest topology with this property is the Alexandrov topology for which continuity coincides with monotonicity. A very reasonable choice of topology induced by a partial order might be the Scott topology which leads to the concept of a Scott continuity: for reasonable posets a map is Scott-continuous when it preserves directed suprema. You may entertain yourself by figuring out whether there is a topology for which the continuous maps are those that preserve suprema and infima.

I cannot tell you which topology is the right one for you. That depends on what you are doing. I hope it is at least clear that the question should be framed in the context of algebra and category theory (morphisms preserve structure) and that continuity is about topology (so we should topologize partial orders).

Answer 2

They are not equivalent even for complete lattices. For example, consider the lattice (actually, frame) $\textrm{Ouv}(X)$ of open sets of a topological space $X$. Given a continuous map $f : X \to Y$, taking inverse images gives a lattice homomorphism $f^{-1} : \textrm{Ouv}(Y) \to \textrm{Ouv}(X)$, and this is guaranteed to preserve infinite joins. However, infinite meets need not be preserved: for example, consider the "identity" map $f : \mathbb{R}^\textrm{disc} \to \mathbb{R}$, where $\mathbb{R}^\textrm{disc}$ is $\mathbb{R}$ considered as a discrete topological space. Since $\mathbb{R}$ is Hausdorff, $\lbrace 0 \rbrace$ is the intersection of all the open neighbourhoods of $0$, and hence the meet in $\textrm{Ouv}(\mathbb{R})$ of all open neighbourhoods of $0$ is $\emptyset$. But the meet in $\textrm{Ouv}(\mathbb{R}^\textrm{disc}) = \mathscr{P}(\mathbb{R})$ of all open neighbourhoods of $0$ is $\lbrace 0 \rbrace$, so $f^{-1} : \textrm{Ouv}(\mathbb{R}) \to \textrm{Ouv}(\mathbb{R}^\textrm{disc})$ indeed fails to preserve infinite meets.