We can define a limit of a sequence of points in a locale in the usual way: $x$ is a limit of $\{ x\_i \}\_{i \in \mathbb{N}}$ if, for every open $U$ containing $x$, there exists $N$ such that $x\_n$ belongs to $U$ for every $n > N$. Now, let's say that we have a sequence of localic maps $\{ f\_i : X \to Y \}\_{i \in \mathbb{N}}$. Can we define a limit of such a sequence? For topological spaces, we can use pointwise convergence. Of course, we can do the same thing for locales, but it seems less natural and I don't think we can do much with such a definition.


If $X$ is exponential, then we can consider $f\_i$ as points in the locale $Y^X$, so we can apply the definition that I gave before. The problem here is that the definition of $Y^X$ is rather complicated and I'm not sure how to prove anything about such a notion of convergence (like uniqueness) or how to prove that any given specific sequence of maps converges to something. So, is this definition a reasonable one or is there some other definition of convergence for localic maps? It's OK to assume that $X$ and $Y$ satisfy some properties like Hausdorffness or metrizability.



Convergence of localic maps
There is a pretty good notion of convergence of maps of locales, though I have never seen anything in the literature about it (maybe I should write something about it ?).


A map of locale $f:X \to Y$ can be thought of as a point of $Y$ in the internal logic of the sheaf topos $Sh(X)$. And a sequence/net of such maps $(f\_i)\_{i\in I}$ can be thought of as an $I$-indexed sequence of points of $Y$ in the internal logic of $Sh(X)$.


**Definition 1:** I say that the sequence/net $(f\_i)\_{i\in I}$ converge to $f$ if the statement "$(f\_i)\_{i\in I}$ converge de $f$ as points of $Y$" holds in the internal logic of the topos $Sh(X)$.


If you are fluent in internal/external translation, it is not too hard to write down explicitly what this means:


**Definition 2:** $(f\_i)\_{i\in I}$ converge to $f$, if for each open $U \in \mathcal{O}(Y)$ there exists a covering $(V\_j)$ of $f^\*(U)$ and for each $j$ an index $i\_j \in I$ such that for all $i > i\_j$, $V\_j \subset f\_i^\*(U)$.


And you can use definition 2 as your starting point if you don't like internal logic. Note that definition $2$ can also be written in the more compact, though harder to read way:


**Definition 3:** $(f\_i)\_{i \in I} \to f$ if for all $U \in \mathcal{O}(Y)$ one has


$$ f^\* U \subset \bigcup\_{i\in I} \left( \bigcap\_{j>i} f\_j^\*(U) \right) $$


It is easy to see that if $X$ is compact then this is the same as uniform convergence and if $X$ is locally compact then this corresponds to uniform convergence on all compacts.


It seems the analogue notion for topological spaces has been studied under the name **"continuous convergence"**. I had mention this a few years ago in [another MO answer](https://mathoverflow.net/questions/187193/between-compact-and-locally-uniform-what-is-the-name-of-this-convergence/187474#187474). (I had given a reference in this other answer, but I honestly don't remember it).


There is also another point of view on this notion of convergence which is quite convenient. I'll restrict to $I = \mathbb{N}$ for simplicity. I'm writting $\overline{\mathbb{N}}$ for the topological space $\mathbb{N} \cup \{ \infty \}$ with the topology whose open are all subsets of $\mathbb{N}$ and all subsets containing $\infty$ and all large enough integer. So a continuous map $f:\overline{\mathbb{N}} \to X$ is the same as a sequence of points $(f(n))\_{n\in \mathbb{N}}$ that converge to $f(\infty)$. Then we have:


**Proposition:** A sequence of maps of locale $(f\_n:X \to Y)\_{n \in \mathbb{N}}$ converge to $f :X \to Y$ in the sense above if and only if there exists a map of locale $F: X \times \overline{\mathbb{N}} \to Y$ such that $F$ restricted to $X \times \{n\}$ is $f\_n$ and $F$ restricted to $X \times \{\infty\}$ if $f$.


I feel like writting the details of the proof would be a bit too long for an MO answer, but a key point if you are trying to prove it is that one can describe explicitly the locale $X \times \overline{\mathbb{N}}$ : an open is a collection $(U\_i)\_{i \in \overline{\mathbb{N}}}$ such that:


$$U\_\infty \subset \bigcup\_{n \in \mathbb{N}} \left( \bigcap\_{m >n} U\_m \right) $$


In particular, if $X$ is exponentiable, then as a morphism $X \times \overline{\mathbb{N}} \to Y$ is the same as a morphism $\overline{\mathbb{N}} \to Y^X$, one deduces that:


**Corollary:** If $X$ is exponentiable, then a sequence of function $f\_n:X \to Y$ converge to $f:X \to Y$ in the sense above if and only if $f\_n$ converge to $f$ as points of $Y^X$.


