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

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

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

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

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

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

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

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