Continuous extensions from dense subspaces (pointfree)
It’s a well-known fact of classical topology that continuous maps (onto Hausdorff spaces) are entirely determined by their behavior on dense subspaces of the input domain. For instance, a continuous function is entirely determined by its values on rational inputs (since is dense in ).
This fact becomes much more useful in pointfree topology, where there are many more dense subspaces than in classical topology. Some dense subspaces might not even have any points. For instance, any probability distribution over a space is entirely specified by its restriction to the subspace , which is the smallest subspace of that has probability 1 under [Simpson 2011]. If is a non-atomic measure, then doesn’t have any points. Still, will be dense if the support of is all of . For instance, if is a normal distribution, then will be dense but have no points.
Using these random subspaces can help to clean up classical probability theory. For instance, in classical probability theory, disintegrations of a probability distribution are only unique up to probability-1 subsets. Given a probability distribution over a product space , let be its marginal distribution over . Then is a disintegration of if Notice that isn’t necessarily unique. Two disintegrations may differ on any null sets. What this means is that a disintegration isn’t guaranteed to provide actually useful information on all its arguments; sometimes its results can be completely arbitrary. For instance, if is a Dirac delta distribution on some point , then can be arbitrary on parts of which are away from , so it would be silly to try to impute meaning from ’s behavior away from . This dilemma has been discussed [Shan & Ramsey 2016].
Only by switching to pointfree topology can we solve the probability-1 dilemma. We simply restrict to , since has the special property that all of its “non-empty” parts have non-zero measure. Therefore, disintegrations of must in fact be unique.
Still, this might be disappointing if has no points, as sometimes we might be interested in inspecting the value of the disintegration at points of the original space . Therefore, we might be interested to know when we can continuously extend to a map , and when must such an extension itself be unique?
This note focuses on the second question, giving it a slightly more general phrasing. Given a function from a subspace of , when is uniqueness guaranteed for continuous extensions ? We will not worry about existence for now.
We will show that if is dense in and if is Hausdorff, then continuous extensions (if they exist) must be unique. Recall that is dense iff the nucleus that presents it has . is Hausdorff iff the diagonal relation on is closed.
Since is Hausdorff, there is an open subspace and accordingly, its complement, the subspace of where and agree, is closed and is the equalizer .
Every closed subspace is co-classified by an open set such that its nucleus is defined by
Since and agree on the dense subspace , it must be that is a subspace of , meaning that is also dense, so that its nucleus satisfies . But since is also closed, it has for some , and therefore it must be that , meaning that is the identity, and that in fact is the entire space , which in turn yields the desired result that and agree everywhere.
Here is a quick counterexample to show why must be Hausdorff. Suppose that both and are the Sierpínski space , and that the dense subspace of is the one which includes only the point of (the “open” point). Then the function which maps to can be extended both the identity map as well as the constant map, which are distinct, even though both are continuous extensions of .