# 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 $f : \mathbb{R} \to \mathbb{R}$ 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 $f$.