Relatively compact subsets in $\mathsf{Top}(X, Y)$ for the product topology (Exercise 2.19 in “Topology: A Categorical Approach”) I am trying to make sense of Exercise 2.19 in Topology: A Categorical Approach by Tai-Danae Bradley, Tyler Bryson and John Terilla.

Exercise 2.19.
If $X$ is any set and $Y$ is Hausdorff, then a subset $\newcommand{\Top}{\mathsf{Top}} A ⊆ \Top(X, Y)$ has compact closure in the product topology if and only if for each $x ∈ A$, the set $A_x = \{ f(x) \mid f ∈ A \}$ has compact closure in $Y$.

My question is essentially the following:

Question.
What is ‘the correct’ formulation/statement of this exercise?

My problems with the given formulation are twofold:

*

*The notation $\Top(X, Y)$ stands for the set of continuous maps from $X$ to $Y$.
This doesn’t make sense if $X$ is only a set.
So either we should require $X$ to be a topological space, or we need to use $\newcommand{\Map}{\mathsf{Map}} \Map(X, Y)$ instead of $\Top(X, Y)$.


*When referring to “the product topology”, I assume that we identify $\Map(X, Y)$ with the product $\prod_{x ∈ X} Y$, so that the projection onto the $x$-th factor correspond to the evaluation at $x$.
But it is not clear to me if we’re talking about the closure of $A$ in $\Top(X, Y)$ (endowed with the subspace topology) or its closure in $\Map(X, Y)$.
I have so far not found an interpretation of the exercise which is both correct and also uses that $A$ consists of continuous maps, instead of just arbitrary ones.
More explicitly, I have tried the following so far:

*

*If we let $X$ be a topological space and take the closure in $\Top(X, Y)$, then the equivalence seems to be wrong.¹


*If we let $X$ be a topological space and take the closure in $\Map(X, Y)$, then the equivalence seems to be true;
but not only for subsets of $\Top(X, Y)$, but for arbitrary subsets of $\Map(X, Y)$.


*If we let $X$ be just a set and work with $\Map(X, Y)$ instead of $\Top(X, Y)$, then the equivalence seems to be true.
This is basically equivalent to the previous point by regarding $X$ as a discrete space.
While the last two formulations seem to be correct, they are not really about (continuous) functions $X \to Y$ anymore, but instead about arbitrary products $∏_α Y_α$ of Hausdorff spaces.
This seems to betray the way the exercise is presented, which may or may not be intended.

¹ If we take $Y$ to also be compact, then the second condition is true for every choice of $A$, in particular for $A = \Top(X, Y)$.
Thus, $\Top(X, Y)$ would be compact, and hence closed in $\Map(X, Y)$ (which is again Hausdorff).
This would then entail that pointwise limits of continuous maps from $X$ to $Y$ are again continuous, which has the usual counterexamples.
 A: One fruitful generalization of the exercise is to characterize subsets of $\mathsf{Top}(X,Y)$ that are relatively compact for the coarsest topology for which the two-variable evaluation function $\mathsf{Top}(X,Y)\times X\to Y$ is continuous. According to section XVI.6 in the book Convergence Foundations of Topology by Szymon Dolecki and Frederic Mynard, these kind of results are called theorems of Ascoli-Arzela type.

The relationship of such results with the exercise is as follows.
The topology of pointwise convergence on $\mathsf{Maps}(X,Y)$ is the coarsest one for which each evaluation $\mathsf{Maps}(X,Y)\times\{x\}\to Y$ at a point $x\in X$ is continuous. What is happening in the case where $X$ is discrete is that in fact $\mathsf{Maps}(X,Y)=\mathsf{Top}(X,Y)$ and the two-variable evaluation function $\mathsf{Top}(X,Y)\times X\to Y$ is continuous.
More generally, the compact-open topology is the coarsest one for which $\mathsf{Top}(X,Y)\times K\to Y$ are continuous for each compact subset $K\subseteq X$. Likewise, $\mathsf{Top}(X,Y)\times X\to Y$ is then continuous for the compact-open topology on $\mathsf{Top}(X,Y)$ when $X$ is locally compact.
Note that if $A\subseteq\mathsf{Maps}(X,Y)$ has $A\times X\to Y$ continuous, then $A$ necessarily consists of continuous functions.

The implication that $A\subseteq\mathsf{Top}(X,Y)$ (or $A\subseteq\mathsf{Maps}(X,Y)$) relatively compact implies each $A_x\subseteq Y$ is relatively compact holds for any topology on $\mathsf{Top}(X,Y)$ (or $\mathsf{Maps}(X,Y)$) for which evaluation at each point is continuous.
If $\mathsf{Top}(X,Y)\times X\to Y$ is continuous, then according to (the proof of) Theorem XVI.6.8. $A$ has to satisfy an additional condition of being evenly continuous.
In general a subset $A\subseteq\mathsf{Maps}(X,Y)$ is evenly continuous if given a filter of functions containing $A$ that converges to a function $f$, and a filter of points of $X$ converging to a point $x$, the filter generated by applying the former to the latter converges to $f(x)$. According to Corollary XI.6.6. of the book, such sets are in fact contained in $\mathsf{Top}(X,Y)$, hence do deserve the name "continuous".
Theorem VI.6.7. then shows that if $Y$ is regular (or more generally if we read the proof, if any filter of functions converging with respect to joint continuity does so to a continuous function), then conversely $A$ is relatively compact in the compact-open topology if each $A_x$ is compact and $A$ is evenly continuos. The more general statement then accounts for the case of $X$ discrete and the topology of pointwise convergence (since all functions are continuous and a set being evenly continuous is vacuous when $X$ is discrete).

Finally, note that Corollary XVI.6.5. asserts that if $A$ is evenly continuous, then its natural convergence structure is the pointwise one. In the topological setting with $X$ locally compact, this means that the closure of an evenly continuous $A$ in the compact-open topology on continuous functions and in the pointwise topology on all functions coincide.
