Continuous extensions from dense subspaces (pointfree)

Posted on November 28, 2016

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 f:f : \mathbb{R} \to \mathbb{R} is entirely determined by its values on rational inputs (since \mathbb{Q} is dense in \mathbb{R}).

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 μ:(A)\mu : \mathcal{R}(A) over a space AA is entirely specified by its restriction to the subspace Ran(μ)\text{Ran}(\mu), which is the smallest subspace of AA that has probability 1 under μ\mu [Simpson 2011]. If μ\mu is a non-atomic measure, then Ran(μ)\text{Ran}(\mu) doesn’t have any points. Still, Ran(μ)\text{Ran}(\mu) will be dense if the support of μ\mu is all of AA. For instance, if μ\mu is a normal distribution, then Ran(μ)\text{Ran}(\mu) 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 μ\mu over a product space A×BA \times B, let ν:(A)\nu : \mathcal{R}(A) be its marginal distribution over AA. Then f:A(B)f : A \to \mathcal{R}(B) is a disintegration of μ\mu if μ=map(λy.(x,y),f(x))dν(x). \mu = \int \text{map}(\lambda y. (x, y), f(x)) d\nu(x). Notice that ff isn’t necessarily unique. Two disintegrations may differ on any null sets. What this means is that a disintegration ff isn’t guaranteed to provide actually useful information on all its arguments; sometimes its results can be completely arbitrary. For instance, if ν\nu is a Dirac delta distribution on some point a:Aa : A, then ff can be arbitrary on parts of AA which are away from aa, so it would be silly to try to impute meaning from ff’s behavior away from aa. This dilemma has been discussed [Shan & Ramsey 2016].

Only by switching to pointfree topology can we solve the probability-1 dilemma. We simply restrict μ:(A×B)\mu : \mathcal{R}(A \times B) to μ:(Ran(ν)×B)\mu' : \mathcal{R}(\text{Ran}(\nu) \times B), since Ran(ν)\text{Ran}(\nu) has the special property that all of its “non-empty” parts have non-zero measure. Therefore, disintegrations of μ\mu' must in fact be unique.

Still, this might be disappointing if Ran(ν)\text{Ran}(\nu) has no points, as sometimes we might be interested in inspecting the value of the disintegration ff at points of the original space AA. Therefore, we might be interested to know when we can continuously extend f:Ran(ν)(B)f : \text{Ran}(\nu) \to \mathcal{R}(B) to a map g:A(B)g : A \to \mathcal{R}(B), 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 f:SBf : S \to B from a subspace SS of AA, when is uniqueness guaranteed for continuous extensions g:ABg : A \to B? We will not worry about existence for now.

We will show that if SS is dense in AA and if BB is Hausdorff, then continuous extensions (if they exist) must be unique. Recall that SS is dense iff the nucleus j:𝒪(A)𝒪(A)j : \mathcal{O}(A) \to \mathcal{O}(A) that presents it has j()=j(\bot) = \bot. BB is Hausdorff iff the diagonal relation on BB is closed.

Since BB is Hausdorff, there is an open subspace β={b:B|g(b)g(b)}, \beta = \{ b : B \ |\ g(b) \neq g'(b) \}, and accordingly, its complement, the subspace of BB where gg and gg' agree, is closed and is the equalizer eq(g,g)\text{eq}(g, g').

Every closed subspace is co-classified by an open set VV such that its nucleus jj is defined by j(U)=UV.j(U) = U \cup V.

Since gg and gg' agree on the dense subspace SS, it must be that SS is a subspace of eq(g,g)\text{eq}(g, g'), meaning that eq(g,g)\text{eq}(g, g') is also dense, so that its nucleus jj satisfies j()=j(\bot) = \bot. But since eq(g,g)\text{eq}(g, g') is also closed, it has j()=Vj(\bot) = \bot \cup V for some VV, and therefore it must be that V=V = \bot, meaning that jj is the identity, and that in fact eq(g,g)\text{eq}(g, g') is the entire space AA, which in turn yields the desired result that gg and gg' agree everywhere.

Here is a quick counterexample to show why BB must be Hausdorff. Suppose that both AA and BB are the Sierpínski space Σ\Sigma, and that the dense subspace SS of Σ\Sigma is the one which includes only the true\text{true} point of Σ\Sigma (the “open” point). Then the function f:SBf : S \to B which maps true\text{true} to true\text{true} can be extended both the identity map as well as the constant true\text{true} map, which are distinct, even though both are continuous extensions of ff.