From discussions I had with Sam, Scott, and Jack:

To solve the problem, it would suffice to find a reflexive domain $X$ with a retract onto $[0, 1]$ .

This is because if you have a reflexive domain $X$ , that is, an $X$ with a continuous surjective map $f :: X \to X^{X}$ , and $A$ is a retract of $X$ , then there's also a continuous surjective map $g :: X \to A^{X}$ .

Proof: If $A$ is a retract of $X$ then we have a retraction $r :: X \to A$ and a section $s :: A \to X$ with $r \circ s = 1_{A}$ . Construct $g (x) := r \circ f (x)$ . To show that $g$ is a surjection consider an arbitrary $q \in A^{X}$ . Thus, $s \circ q :: X \to X$ . Since $f$ is a surjection there must be some $x$ with $f (x) = s \circ q$ . It follows that $g (x) = r \circ f (x) = r \circ s \circ q = q$ . Since $q$ was arbitrary, $g$ is also a surjection.

[-]Stuart_Armstrong9y30

A small note: it's not hard to construct spaces that are a bit too big, or a bit too small (raising the possibility that a true $X$ lies between them).

For instance, if $I$ is the unit interval, then we can map $I$ onto the countable hypercube $I^{ω}$ ( https://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn.E2.80.93Mazurkiewicz_theorem ). Then if we pick an ordering of the dimensions of the hypercube and an ordering of $Q \cap I$ , we can see any element of $I^{ω}$ - hence any element of $I$ - as a function from $Q \cap I$ to $I$ .

Let $C (I)$ be the space of continuous functions $I \to I$ . Then any element of $C (I)$ defines a unique function $Q \cap I \to I$ (the converse is not true - most functions $Q \cap I \to I$ do not correspond to continuous functions $I \to I$ ). Pulling $C (I)$ back to $I$ via $I^{ω}$ we define the set $Y \subset I$ .

Thus $Y$ maps surjectively onto $C (I)$ . However, though $C (I)$ maps into $C (Y)$ by restriction (any function from $I$ is a function from $Y$ ), this map is not onto (for example, there are more continuous functions from $I - {1 / 2}$ than there are from $I$ , because of the potential discontinuity at $1 / 2$ ).

Now, there are elements of $I - Y$ that map to functions in $C (Y)$ but not in $C (I)$ . So there's a hope that there may exist an $X$ with $Y \subset X \subset I$ , $C (I) \subset C (X) \subset C (Y)$ , and $X$ mapping onto $C (X)$ . Basically, as $X$ `gets bigger', its image in $C (Y)$ grows, while $C (X)$ itself shrinks, and hopefully they'll meet.

[-]Donald Hobson7y30

I think you made a mistake here

The Hahn–Mazurkiewicz theorem states that

A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.

I will agree that $I^{w}$ is connected and locally connected. I'm not sure if its second countable. It is not compact.

Just to be clear $I^{w} = {(x_{1}, x_{2}, \dots) \forall i : x_{i} \in [0, 1]]}$ Now let $H_{i} = {x \in I^{w} | x_{i} < 0.6}$ . Clearly each $H_{i}$ is open. Let $G = {x \in I^{w} | \forall i : x_{i} > 0.4}$ And $F = {G, H_{1}, H_{2}, \dots}$ . Now clearly this family covers all of $I^{w}$ . However, remove any $H_{i}$ from $F$ and $x = (i      0.9, 0.9, \dots 0.9, 0.1, 0.9, 0.9, \dots)$ is no longer covered. So $F$ is a family of open sets, which cover $I^{w}$ and don't have any finite subcover.

[-]Stuart_Armstrong7y30

Hum, $I^{ω}$ should be compact by Tychonoff's theorem (see also the Hilbert Cube, which is homeomorphic to $I^{ω}$ ).

For your proof, I think that $G$ is not open in the product topology. The product topology is the coarsest topology where all the projection maps are continuous.

To make all the projection maps continuous we need all sets in $S$ to be open, where we define $σ \in S$ iff there exists an $i$ , such that $p_{i} (σ)$ is open in $[0, 1]$ and $σ = {x = (x_{1}, x_{2}, \dots) | x_{i} \in p_{i} (σ), 0 \leq x_{j} \leq 1 for i \neq j}$ .

Let $S^{'}$ be the set of finite intersection of these sets. For any $σ^{'} \in S^{'}$ , there exists a finite set $N_{σ^{'}} \subset N$ such that if $x \in σ^{'}$ and $y_{i} = x_{i}$ for $i \in N_{σ^{'}}$ , then $y \in σ^{'}$ as well.

If we take $S^{''}$ to be the arbitrary union of $S^{'}$ , this condition will be preserved. Thus $G$ is not contained in the arbitrary unions and finite intersections of $S$ , so it seems it is not an open sent.

Also, $I^{ω}$ is second-countable. From the wikipedia article on second-countable:

Any countable product of a second-countable space is second-countable

[-]jessicata9y30

If I were studying this, I would be looking at domain theory, in which (among other things) there has been found a topological space $X$ homeomorphic with $X^{X}$ . The page I linked links to some notes at the bottom. (h/t Qiaochu for pointing out domain theory)

[-]Rupert5y*20

I have just now submitted an attempted solution to this problem to "Geometry and Topology". I claim that the space $X$ you are looking for is $2^{ω_{1}}$ ( $ω_{1}$ being the least uncountable cardinal) with the ``generalised Cantor space topology", that is for each countable well-ordered bit-string $b$ you have a basic open set consisting of all bit-strings of length $ω_{1}$ with $b$ as an initial fragment. Since this topological space has quite a large cardinality I'm somewhat unclear whether this is helpful for your proposed application and would need to think about it more. (Matthew Barnett just now directed me to this post of yours.) I sent you an early draft of my paper, which argues the point in detail, on FB Messenger, and can send the latest version to you if you wish.

[-]Rupert5y10

Let $A^{'}$ be $2^{ω_{1}}$ with generalised Cantor space topology, and $A^{''}$ be $2^{ω_{1}}$ with product topology, $X$ a closed disc in a finite-dimensional Euclidean space. Then there is a continuous surjection $A^{'} \to X^{A^{''}}$ . I don't know how to show that there is a topological space $A$ with carrier set $2^{ω_{1}}$ and a continuous surjection $A \to X^{A}$ . Thanks to Alex Mennen for pointing out the problem.

[-]Rupert5y*10

However, because topology on $A^{'}$ is finer than topology on $A^{''}$ here, this still shows how the proof of the Lawvere fixed point theorem can be applied here to give Brouwer fixed point theorem as corollary, which could conceivably be a publishable result (see what "Geometry and Topology" think about that), and this could still be sorta kinda maybe relevant to Scott's original motivation for looking at the problem (if you're okay with working with two different topologies on the space of agents, one finer than the other). But this is a very big space of agents you're talking about here.

Correction: need not only that topology on $A^{'}$ is finer than topology on $A^{''}$ , but also, given arbitrary open subset of $X$ , take pre-image under evaluation map in $X^{A^{''}} \times A^{''}$ , projection onto first factor and then pre-image of that under the continuous surjection $A^{'} \to X^{A^{''}}$ , it needs to be shown that this set is open in both topologies. I believe that this can indeed be done for an appropriate class of spaces $X$ for the pair of topologies in question.

[-]Rupert5y10

When I look at my post the LaTeX code isn't formatting properly; if anyone can let me know how to fix that.

[-]Ben Pace5y10

I fixed it. In our editor, use cmd-4/ctrl-4 to do LaTex, not dollar signs. (The thing you did would work in the markdown editor – you can go into settings to change to that editor if you'd like.)

[-]Stuart_Armstrong7y10

EDIT: This idea is wrong: https://www.lesswrong.com/posts/eqi83c2nNSX7TFSfW/no-surjection-onto-function-space-for-manifold-x

Ok, here's an idea for constructing such a map, with a few key details left unproven; let me know if people see any immediate flaws in the approach, before I spend time filling in the holes.

Let $X$ be a countable collection of open intervals (eg $X = {x \in R, x \notin N}$ ), given the usual topology. Let $I = [0, 1]$ be the closed unit interval, and $C (X, I)$ the set of continuous functions from $X$ to $I$ . Give $C (X, I)$ the compact-open topology.

By the properties of the compact-open topology, since $I$ is $T_{3.5}$ (Tychonoff), then so is $C (X, I)$ . I'm hoping that the proof can be extended, at least in this case, to show that $C (X, I)$ is $T_{4}$ (normal Haussdorff).

It seems clear that $C (X, I)$ is second-countable: let $V (K, U)$ consist of all functions that map $K$ into $U$ , where $K \subset X$ is the intersection of $X$ with a closed interval with rational endpoints, and $U \subset I$ is an open interval with rational endpoints. The set of all such $V (K, U)$ is countable, and forms a subbasis of $C (X, I)$ . A countable subbasis means a countable basis, as the set of finite subsets of countable set, is itself countable.

If $C (X, I)$ is $T_{4}$ and second countable, then it is homeomorphic to a subset of the Hilbert Cube. To simplify notation, we will identify $C (X, I)$ with its image in the Hilbert Cube.

Take the closure $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ C (X, I)$ of $C (X, I)$ within the Hilbert Cube. This closure is compact and second countable (since the Hilbert Cube itself is both). It seems clear that $C (X, I)$ is connected and locally connected; connected will extend to the closure, we'll need to prove that locally connected does as well.

Then we can apply the Hahn-Mazurkiewicz theorem:

A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.

So there is a continuous surjection $ϕ : I \to ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ C (X, I)$ . Pull back $C (X, I)$ , defining $ϕ^{- 1} (C (X, I)) \subset I$ . If $C (X, I)$ is open in $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ C (X, I)$ (with the subspace topology), then $ϕ^{- 1} (C (X, I))$ is open in $I$ . Even if $ϕ^{- 1} (C (X, I))$ is not open, we can hope that, at worst, it consists of a countable collection of points, open intervals, half-closed intervals, and closed intervals (this is not a general property of subsets of the interval, cf the Cantor set, but it feels very likely that it will apply here).

In that case, these is a continuous surjection $s$ from $X$ to $ϕ^{- 1} (C (X, I))$ , mapping each open interval to one of the points or intervals ("folding" over the ends when mapping to those with closed end-points).

Then $ϕ \circ s : X \to C (X, I)$ is the continuous surjection we are looking for.

Note: I'm thinking now that $C (X, I)$ might not be connected, but this would not be a problem as long as it has a countable number of connected components.

[-]Dacyn8y10

I am going to take some license with your question because I think you are asking the wrong question. Arbitrary topological spaces and abstract continuity are rarely the right notions in real-world situations. Rather, uniform continuity on bounded sets usually better corresponds to the intuitive notion of "a small change in input produces a small change in output".

Thus, suppose that $X$ is a complete separable metric space and that $f : X \times X \to [0, 1]$ is uniformly continuous on bounded sets. Then we can show that there exists a function $g : X \to [0, 1]$ which is uniformly continuous on bounded sets but not a fiber of $f$ (i.e. there is no $x$ such that $f (x, y) = g (y)$ for all $y$ ). Indeed, consider two cases:

$X$ is uniformly discrete. Then every map from $X$ to $[0, 1]$ is uniformly continuous, so we get a contradiction from cardinality considerations.
$X$ is not uniformly discrete. Then for each n, since f is uniformly continuous on $B (0, n)$ it has a modulus of continuity on this set, i.e. a continuous increasing function $h_{n} : (0, \infty) \to (0, \infty)$ such that $d (f (x), f (y)) < h_{n} (d (x, y))$ for all $x, y \in B (0, n) \subseteq X \times X$ . Since $X$ is not uniformly discrete, there is a function $g : X \to [0, 1]$ such that for infinitely many $n$ , there exist $x, y$ such that $d (g (x), g (y)) > h_{n} (d (x, y))$ . (To construct it, take pairs $(x_{n}, y_{n})$ with $h_{n} (d (x_{n}, y_{n})) < 2^{- n}$ , extract a subsequence that behaves geometrically nicely, and then find a function $g$ such that $d (g (x_{n}), g (y_{n})) > 2^{- n}$ for all $n$ in the subsequence.) Clearly, $g$ cannot be a fiber of $f$ .

[-]AlexMennen8y10

$g$ can be a fiber of $f$ , since for each $n$ , $x_{n}$ and $y_{n}$ could be distance greater than $n$ from the basepoint.

Example: let $X := {x_{n}, y_{n} | n \in N}$ , with $d (x_{n}, x_{m}) = d (y_{n}, y_{m}) = d (x_{n}, y_{m}) = 2 | n - m |$ for $n \neq m$ and $d (x_{n}, y_{n}) = 6^{- n}$ . Let $x_{0}$ be the basepoint (so that $(x_{0}, x_{0})$ is the point you were calling " $0$ "). Let $g (x_{n}) := 0$ , $g (y_{n}) := 1$ , $f (z, w) := g (z)$ , and $h_{n} (r) := 3^{n} r$ .

I also don't see how to even construct the function $g$ , or, relatedly, what you mean by "geometrically nicely", but I guess it doesn't matter.

Also, I'm not convinced that metric spaces with uniform continuity on bounded subsets is a better framework than topological spaces with continuity.

[-]AlexMennen8y00

This is intended as a reply to David Simmons's comment, which for some reason I can't reply to directly.

In the new version of your proof, how do we know $Y_{k}$ isn't too close to $X_{l}$ for some $l > k$ ? And how do we know that $g$ is uniformly continuous on bounded subsets?

About continuity versus uniform continuity on bounded sets:

It seems to me that your point 1 is just a pithier version of your point 4, and that these points support paying attention to uniform continuity, rather than uniform continuity restricted to bounded sets. This version of the problem seems like it would be a less messy version of the "fixed modulus of continuity" version of the problem you mentioned (which I did not understand your solution to, but will look again later).

I'm not sure what you're getting at about singularities in point 3. I wouldn't have asked why you were considering uniform continuity on bounded sets instead of uniform continuity away from singularities (in fact, I don't know what that means). I would ask, though, why uniform continuity on bounded sets instead of uniform continuity on compact sets? As you point out, the latter is the same as continuity.

Your point 2 is completely wrong, and in fact this is the primary reason I was convinced that continuity is a better thing to pay attention to than uniform continuity on bounded sets. The type of object you are describing is an effective Polish space that remembers its metric. Typically, descriptive set theorists forget the metric, and the isomorphisms between Polish spaces are homeomorphisms (and the isomorphisms between effective Polish spaces are computable homeomorphisms). Changing the metric in a computably homeomorphic way does not change what can be done when points are represented as descending chains of basic open sets with singleton intersection. So the thing you described was really topological rather than metric in nature, even though it involves introducing a metric in the setup. I am not aware of any notions of computability in metric spaces in which the metric matters in the way you are suggesting. It is not true that uniform continuity gives you an algorithm for computing the function. As a counterexample, let $a$ be an uncomputable number, and let $f : R \to R$ be given by $f (x) = a$ for every $x$ . $f$ is clearly uniformly continuous, but not computable. It is also not true that uniform continuity is necessary in order for the function to be computable. For instance, $sin (1 / x)$ is computable on $(0, 1]$ . Of course, $(0, 1]$ is not complete, but for another example, consider an effective infinite-dimensional Hilbert space, and let ${e_{n} | n \in N}$ be an effective orthonormal basis. Let $f_{n} (x) := max (0, \frac{1}{2} - ∥ x - e_{n} ∥)$ . This sequence of functions is computable, they have disjoint support, and for any point, a sufficiently small neighborhood around it will be disjoint from the supports of all but at most one of these functions. Thus $f (x) := \sum_{n} n f_{n} (x)$ is computable, but of course it is not uniformly continuous on the unit ball, which is bounded. However, it is true that every computable function is continuous, and conversely, every continuous function is computable with respect to some oracle. Of course, we really want computability, not computability with respect to some oracle, but this still seems to show that continuity is at least a good metaphor for computability, whereas uniform continuity on bounded sets doesn't seem so to me.

Of course, all this about continuity as a metaphor for computability makes the most sense in the context of Polish spaces, and we can only talk about actual computability in the context of effective Polish spaces. Scott's problem involves a space $X$ and the exponential $[0, 1]^{X}$ . If $X$ is a locally compact Polish space, then $[0, 1]^{X}$ is also Polish. I think that this might be necessary (that is, if $X$ and $[0, 1]^{X}$ are Polish, then $X$ is locally compact), although I'm not sure. If so, and if your proof is correct, it seems plausible that your proof could be adapted to show that there is no locally compact Polish space with the property that Scott was looking for, and that would show that there is no solution to the problem in which $X$ and $[0, 1]^{X}$ are both Polish spaces, and hence no computable solution, if computability is formalized as in effective descriptive set theory.

[-]Dacyn8y10

I don't know why my comment doesn't have a reply button. Maybe it is related to the fact that my comment shows up as "deleted" when I am not logged in.

Sorry, I seem to be getting a little lazy with these proofs. Hopefully I haven't missed anything this time.

New proof: ... We can extract a subsequence $(n_{k})$ such that if $X_{k} = x_{n_{k}}$ and $Y_{k} = y_{n_{k}}$ , then $d (X_{k + 1}, Y_{k + 1}) \leq (1 / 6) d (X_{k}, Y_{k})$ for all $k$ , and for all $k$ and $l > k$ , either (A) $d (X_{k}, Y_{l}) \geq (1 / 3) d (X_{k}, Y_{k})$ and $d (X_{k}, X_{l}) \geq (1 / 3) d (X_{k}, Y_{k})$ or (B) $d (Y_{k}, X_{l}) \geq (1 / 3) d (X_{k}, Y_{k})$ and $d (Y_{k}, Y_{l}) \geq (1 / 3) d (X_{k}, Y_{k})$ . By extracting a further subsequence we can assume that which of (A) or (B) holds depends only on $k$ and not on $l$ . By swapping $X_{k}$ and $Y_{k}$ if necessary we can assume that case (A) always holds.

Lemma: For each $z$ there is at most one $k$ such that $d (z, X_{k}) \leq (1 / 6) d (X_{k}, Y_{k})$ .

Proof: Suppose $d (z, X_{k}) \leq (1 / 6) d (X_{k}, Y_{k})$ and $d (z, X_{l}) \leq (1 / 6) d (X_{l}, Y_{l})$ , with $k < l$ . Then $d (X_{k}, X_{l}) < (1 / 3) d (X_{k}, Y_{k})$ , a contradiction.

It follows that by extracting a further subsequence we can assume that $d (Y_{k}, X_{l}) \geq (1 / 6) d (X_{l}, Y_{l})$ for all $l > k$ .

Now let $j : [0, \infty) \to [0, \infty)$ be an increasing uniformly continuous function such that $j (0) = 0$ and $j ((1 / 6) d (X_{k}, Y_{k})) > 2^{- n_{k}}$ for all $k$ . Finally, let $g (x) = {inf}_{k} j (d (x, Y_{k}))$ . Then for all $k$ we have $g (Y_{k}) = 0$ . On the other hand, for all $k < l$ we have $d (X_{k}, Y_{l}) \geq (1 / 3) d (X_{k}, Y_{k})$ , for $k = l$ we have $d (X_{k}, Y_{l}) = d (X_{k}, Y_{k})$ , and for $k > l$ we have $d (X_{k}, Y_{l}) \geq (1 / 6) d (X_{k}, Y_{k})$ . Thus $g (X_{k}) = {inf}_{l} j (d (X_{k}, Y_{l})) \geq j ((1 / 6) d (X_{k}, Y_{k})) > 2^{- n_{k}}$ . Clearly, $g$ cannot be a fiber of $f$ . Moreover, since the functions $j$ and $x \mapsto {inf}_{k} d (x, Y_{k})$ are both uniformly continuous, so is $g$ .

Regarding your responses to my points:

I guess I don't disagree with what you write regarding my points 1 and 4.

It seems to be harder than expected to explain my intuitions regarding singularities in point 3. Basically, I think the reasons that abstract continuity came to be considered standard are mostly the fact that in concrete applications you have to worry about singularities, and this makes uniform continuity a little more technically annoying. But in the kind of problem we are considering here, it seems that continuity is really more annoying to deal with than uniform continuity, with little added benefit. I guess it also depends on what kinds of functions you expect to actually come up, which is a heuristic judgement. Anyway it might not be productive to continue this line of reasoning further as maybe our disagreements just come down to intuitions.

Regarding my point 2, I wasn't very clear when I said that uniform continuity gives you an algorithm, what I meant was that if you have an algorithm for computing the images of points in the dense sequence and for computing the modulus of continuity function, then uniform continuity gives you an algorithm. The function $x \mapsto sin (1 / x)$ would be the kind of thing I would handle with uniform continuity away from singularities (to fix a definition for this, let us say that you are uniformly continuous away from singularities if you are uniformly continuous on sets of the form $B (0, n) ∖ B (S, 1 / n)$ , where $S$ is some set of singularities).

In your definition of $f_{n}$ , I think you mean to write max instead of min. But I see your point, though the example seems a little pathological to me.

Anyway, it seems that you agree that it makes sense to restrict to Polish spaces based on computability considerations, which is part of what I was trying to say in 2.

If you have a locally compact Polish space, then you can find a metric with respect to which the space is proper (i.e. bounded subsets are compact): let $d^{'} (x, y) = d (x, y) + | f (x) - f (y) |$ , where $f (x) = 1 / sup {r : B (x, r)$ is compact $}$ . With respect to this metric, continuity is the same as uniform continuity on bounded sets, so my proof should work then.

Proposition: Let $X$ be a Polish space that is not locally compact. Then $[0, 1]^{X}$ (with the compact-open topology) is not first countable.

Proof: Suppose otherwise. Then the function $f_{0} \equiv 0$ has a countable neighborhood basis of sets of the form $F (K_{n}, U_{n}) = {f : f (K_{n}) \subseteq U_{n}}$ where $K_{n} \subseteq X$ is compact and $U_{n} \subseteq [0, 1]$ is open. Since $X$ is not locally compact, there exists a point $x$ such that $B (x, r)$ is not compact for any $x$ . For each $n$ , we can choose $x_{n} \in B (x, 1 / n) ∖ ⋃_{i \leq n} K_{i}$ . Let $K = {x_{n} : n} \cup {x}$ , and note that $K$ is compact. Then $F (K, [0, 1 / 2))$ is a neighborhood of $f_{0}$ . But then $F (K, [0, 1 / 2)) \supseteq ⋂_{i \leq n} F (K_{i}, U_{i})$ for some $n$ . This contradicts the fact that $x_{n} \in K ∖ ⋃_{i \leq n} K_{i}$ , since we can find a bump function which is $0$ on $⋃_{i \leq n} K_{i}$ but $1$ at $x$ .

It does still seem to me that most of the useful intuition comes from point 4 of my previous comment, though.

[-]AlexMennen8y00

It appears that comments from new users are collapsed by default, and cannot be replied to without a "Like". These seem like bad features.

Your proof that there's no uniformly continuous on bounded sets function $f : X \times X \to [0, 1]$ admitting all uniformly continuous on bounded sets functions $X \to [0, 1]$ as fibers looks correct now. It also looks like it can be easily adapted to show that there is no uniformly continuous $f : X \times X \to [0, 1]$ admitting all uniformly continuous functions $X \to [0, 1]$ as fibers. Come to think of it, your proof works for arbitrary metric spaces $X$ , not just complete separable metric spaces, though those are nicer.

I see what you mean now about uniform continuity giving you an algorithm, but I still don't think that's specific to uniform continuity in an important way. After all, if you have an algorithm for computing images of points in the countable dense set, and a computable "local modulus of continuity" in the sense of a computable function $h : X \times [0, \infty) \to [0, \infty)$ with $h (x, 0) = 0$ and $d (x, y) < r ⟹ | f (y) - f (x) | < h (x, r)$ , then $f$ is computable, and this does not require $f$ to be uniformly continuous. Although I suppose you could object that this is a bit circular, in that I'm assuming the "local modulus of continuity" is computable only in the standard sense, which does not require uniform continuity.

I'm not sure why you would allow singularities at some points (presumably a uniformly discrete set, or something like that) while still insisting on uniform continuity elsewhere. It still seems to me that the arguments for uniform continuity rather than continuity all point to wanting uniform continuity entirely, rather than some sense of local uniform continuity in most places.

Thanks for pointing out the error in my definition of $f_{n}$ ; I've fixed it.

In your argument that locally compact Polish spaces can be given metrics with respect to which they are proper, it isn't true that $d^{'}$ is necessarily a proper metric. For instance, consider a countably infinite set with $d (x, y) = 1$ for $x \neq y$ . This is a locally compact Polish space, but $f (x) = 1$ for every $x$ , so $d^{'} = d$ , and the space is not proper.

Your last proposition looks correct (though with a typo: last $\subseteq$ in the proof should be $\supseteq$ ). However, if $X$ is not locally compact, then the compact-open topology isn't necessarily the right topology to consider on $[0, 1]^{X}$ . We want a topology making $[0, 1]^{X}$ into an exponential object, and it isn't clear that such a topology even exists, or that it is the compact-open topology if it does exist (though it must be a refinement of the compact-open topology if it does exist). Maybe asking about non-locally compact Polish spaces $X$ with a Polish exponential space $[0, 1]^{X}$ is a kind of weird question, though, and if we're even considering non-locally compact Polish spaces, we should turn to the version of the question where we just want a continuous function $X \times X \to [0, 1]$ admitting all continuous functions $X \to [0, 1]$ as fibers.

[-]Dacyn8y10

I will have to think more about the issue of continuity vs uniform continuity. I suppose my last remaining argument would be the fact that Bishop--Bridges' classic book on constructive analysis uses uniform continuity on bounded sets rather than continuity, which suggests that it is probably better for constructive analysis at least. But maybe they did not analyze the issue carefully enough, or maybe the relevant issues here are for some reason different.

To fix the argument that every locally compact Polish space admits a proper metric, let $f$ be as before and let $F (x, y) = \infty$ if $d (x, y) \geq f (x)$ and $F (x, y) = f (x) / [f (x) - d (x, y)]$ if $d (x, y) < f (x)$ . Next, let $g (y) = {min}_{n} [n + F (x_{n}, y)]$ , where $(x_{n})$ is a countable dense sequence. Then $g$ is continuous and everywhere finite. Moreover, if $S = g^{- 1} ([0, N])$ , then $S \subseteq ⋃_{n \leq N} B (x_{n}, (1 - 1 / N) f (x_{n}))$ and thus $S$ is compact. It follows that the metric $d^{'} (x, y) = d (x, y) + | g (y) - g (x) |$ is proper.

Anyway I have fixed the typo in my previous post.

[-]AlexMennen8y00

Hm, perhaps I should figure out what the significance of uniform continuity on bounded sets is in constructive analysis before dismissing it, even though I don't see the appeal myself, since constructive analysis is not a field I know much about, but could potentially be relevant here.

$f$ is the reciprocal of what it was before, but yes, this looks good. I am happy with this proof.

[-]Scott Garrabrant8y10

I give a stronger version of this problem here.

[-]Marcello9y10

"Self-Reference and Fixed Points: A Discussion and an Extension of Lawvere's Theorem" by Jorge Soto-Andrade and Francisco J. Varela seems like a potentially relevant result. In particular, they prove a converse Lawvere result in the category of posets (though they mention doing this for $[0, 1]$ in an unsolved problem.) I'm currently reading through this and related papers with an eye to adapting their construction to $[0, 1]$ (I think you can't just use it straight-forwardly because even though you can build a reflexive domain with a retract to an arbitrary poset, the paper uses a different notion of continuity for posets.)

[-]Stuart_Armstrong9y10

Can you argue that $X$ must have a semi-metric compatible with the topology by using $d (x, y) = s u p_{z \in X} | h (x, z) - h (y, z) |$ ?

I'm wondering if you can generalise this to some sort of argument that goes like this. Using X, project down via $π$ from $X^{0} = X$ to $X^{1} = X^{0} / d$ . Let $ϕ$ be our initial surjection; it's now a bijection between $X^{1}$ and maps from $X^{0}$ to $[0, 1]$ .

If the projection is continuous, then every map from $X^{1}$ to $[0, 1]$ lifts to a map from $X^{0}$ to $[0, 1]$ . Restricting to the subset of maps that are lifts like this, and applying $ϕ^{- 1}$ , gives a subset $X^{2} \subset X^{1}$ . We now have a new equivalence relationship, maps from $X^{1}$ that are equal to each other on $X^{2}$ . Project down from $X^{2}$ by this relationship, to generate $X^{3}$ . Continue this transfinitely often (?) to generate a space $X^{'}$ where $ϕ$ is a homeomorphism, and find a contradiction?

This feels dubious, but maybe worth mentioning...

[-]AlexMennen9y20

I haven't checked that argument carefully, but that sounds like it should give you $X^{'}$ with a continuous bijection $ϕ : X^{'} \to [0, 1]^{X^{'}}$ , which might not necessarily be a homeomorphism.

[-]Stuart_Armstrong9y00

Yes, you're right.

[-]AlexMennen8y00

This comment is to explain partial results obtained by David Simmons in this thread, since the results and their proofs are difficult to follow as a result of being distributed across multiple comments, with many errors and corrections. The proofs given here are due to David Simmons.

#Statements

Theorem 1: There is no metric space $X$ and uniformly continuous function $f : X \times X \to [0, 1]$ such that every uniformly continuous function $X \to [0, 1]$ is a fiber of $f$ .

Theorem 2: There is no metric space $X$ and function $f : X \times X \to [0, 1]$ that is uniformly continuous on bounded sets, such that every function $X \to [0, 1]$ which is uniformly continuous on bounded sets is a fiber of $f$ .

Theorem 3: There is no locally compact Polish space $X$ and continuous function $f : X \times X \to [0, 1]$ such that every continuous function $X \to [0, 1]$ is a fiber of $f$ .

Commentary

Theorem 1 says that there is no solution to the version of Scott's problem with continuity in topological spaces replaced with uniform continuity in metric spaces. A plausible reason that this version of the problem might be important is that if an agent has to be able to compute what its policy says to do to any desired precision using an amount of computational resources that does not depend on the input, then its policy must be uniformly continuous. The gist of the proof is that for any uniformly continuous $f : X \times X \to [0, 1]$ , it is possible to construct a uniformly continuous function $g : X \to [0, 1]$ that requires greater precision in its input than $f$ does to determine the output to any desired precision. I suspect it might be possible to adapt the proof to work in uniform spaces rather than just metric spaces, but I have not studied uniform spaces.

Theorem 2 is similar to theorem 1, but with uniform continuity replaced with uniform continuity on bounded subsets. I was not convinced that this is an important notion in its own right, but theorem 2 is useful as a lemma for theorem 3. See the thread under David Simmons's comment for more discussion about what sort of continuity assumption is appropriate. The gist of the proof is to apply the proof of theorem 1 on a bounded set. The function $g$ constructed will be uniformly continuous everywhere, so the proof actually shows a stronger result that unifies both theorems 1 and 2: there is no $f : X \times X \to [0, 1]$ that is uniformly continuous on bounded sets and admits every uniformly continuous function $X \to [0, 1]$ as a fiber.

Theorem 3 almost says that Scott's problem cannot be solved with Polish spaces. It doesn't quite say that, because there are Polish spaces that are not locally compact. However, non-locally-compact Polish spaces are not exponentiable, so in the version of the problem in which we want a surjection $X \to [0, 1]^{X}$ , it isn't even clear whether there exists an exponential object $[0, 1]^{X}$ , which may mean that non-exponentiable spaces are not promising, although I'm not sure. A reason to restrict attention to Polish spaces is that effective Polish spaces provide a topological setting in which there is a good notion of computability, so the nonexistence of a solution in Polish spaces would make it impossible to provide a computable solution in that sense. That said, there may be other notions of computability in topological spaces (domain theory?), which I am unfamiliar with. The gist of the proof is to find a metric in which bounded sets are compact, and apply theorem 2.

#Proofs

Proof of theorem 1: Let $X$ be a metric space, and $f : X \times X \to [0, 1]$ be uniformly continuous. If $X$ is uniformly discrete, then all functions from $X$ are uniformly continuous, so there is a uniformly continuous function $X \to [0, 1]$ that is not a fiber of $f$ by Cantor's theorem.

So assume that $X$ is not uniformly discrete. Let $(x_{n}, y_{n})_{n \in N}$ be such that $\forall k$ $d (x_{k + 1}, y_{k + 1}) \leq \frac{1}{6} d (x_{k}, y_{k})$ . Note that for all $k$ and $ℓ > k$ , either (A) $d (x_{k}, y_{ℓ}) \geq \frac{1}{3} d (x_{k}, y_{k})$ and $d (x_{k}, x_{ℓ}) \geq \frac{1}{3} d (x_{k}, y_{k})$ or (B) $d (y_{k}, y_{ℓ}) \geq \frac{1}{3} d (x_{k}, y_{k})$ and $d (y_{k}, x_{ℓ}) \geq \frac{1}{3} d (x_{k}, y_{k})$ . By extracting a subsequence we can assume that which of (A) or (B) holds depends only on $k$ and not on $ℓ$ . By swapping $x_{k}$ and $y_{k}$ if necessary we can assume that case (A) always holds.

For each $z$ there is at most one $k$ such that $d (z, x_{k}) < \frac{1}{6} d (x_{k}, y_{k})$ , because if $d (z, x_{k}) < \frac{1}{6} d (x_{k}, y_{k})$ and $d (z, x_{ℓ}) < \frac{1}{6} d (x_{ℓ}, y_{ℓ})$ with $ℓ > k$ , then $d (x_{k}, x_{ℓ}) < \frac{1}{3} d (x_{k}, y_{k})$ , a contradiction.

It follows that by extracting a further subsequence we can assume that $d (y_{k}, x_{ℓ}) \geq \frac{1}{6} d (x_{ℓ}, y_{ℓ})$ for all $ℓ > k$ .

Since $f$ is uniformly continuous, there is a function $δ : (0, \infty) \to (0, \infty)$ such that $\forall ε > 0 d ((u, v), (w, z)) < δ (ε) \to | f (u, v) - f (w, z) | < ε$ . By extracting a further subsequence, we can assume that $d (x_{k}, y_{k}) < δ (2^{- k})$ for all $k$ . Let $j : [0, \infty) \to [0, \infty)$ be an increasing uniformly continuous function such that $j (0) = 0$ and $j (\frac{1}{6} d (x_{k}, y_{k})) > 2^{- k}$ for all $k$ . Finally, let $g (z) := {inf}_{k} d (z, y_{k})$ . Then for all $k$ we have $g (y_{k}) = 0$ . On the other hand, for all $ℓ > k$ we have $d (x_{k}, y_{ℓ}) \geq \frac{1}{3} d (x_{k}, y_{k})$ , for $ℓ = k$ we have $d (x_{k}, y_{ℓ}) = d (x_{k}, y_{k})$ , and for $ℓ < k$ we have $d (x_{k}, y_{ℓ}) \geq \frac{1}{6} d (x_{k}, y_{k})$ . Thus $g (x_{k}) = {inf}_{ℓ} j (d (x_{k}, y_{ℓ})) \geq j (\frac{1}{6} d (x_{k}, y_{k})) > 2^{- k}$ . Clearly, $g$ cannot be a fiber of $f$ . Moreover, since the functions $j$ and $z \mapsto {inf}_{k} d (z, y_{k})$ are both uniformly continuous, so is $g$ . $□$

Proof of theorem 2: Let $X$ be a metric space, and $f : X \times X \to [0, 1]$ be uniformly continuous on bounded sets. If all bounded subsets of $X$ are uniformly discrete, then all functions from $X$ are uniformly continuous on bounded sets, so there is a function $X \to [0, 1]$ that is uniformly continuous on bounded sets but not a fiber of $f$ by Cantor's theorem. Otherwise, let $B \subseteq X$ be a bounded set that is not uniformly discrete, take a sequence $(x_{n}, y_{n})_{n \in N}$ in $B$ as in the proof of theorem 1, and a function $δ : (0, \infty) \to (0, \infty)$ such that $\forall ε > 0 d ((u, v), (w, z)) < δ (ε) \to | f (u, v) - f (w, z) | < ε$ for $(u, v), (w, z) \in B \times B$ , and define $j$ and $g$ as in the proof of theorem 1. $g$ is uniformly continuous, but not a fiber of $f$ . $□$

Proof of theorem 3: Let $d$ be a metric for $X$ . Let $f (x) := sup {r | B (x, r) is compact}$ , where $B (x, r)$ is the closed ball around $x$ of radius $r$ . Let $F (x, y) := f (x) / [f (x) - d (x, y)]$ if $f (x) > d (x, y)$ , and $F (x, y) := \infty$ otherwise. Next, let $g (y) := {min}_{n} [n + F (x_{n}, y)]$ , where $(x_{n})_{n \in N}$ enumerates a countable dense set. Then $g$ is continuous and everywhere finite. $g^{- 1} ([0, N]) \subseteq ⋃_{n \leq N} B (x_{n}, (1 - \frac{1}{N}) f (x_{n}))$ , and is thus compact. It follows that with the metric $d^{'} (x, y) := d (x, y) + | g (y) - g (x) |$ , which induces the same topology, bounded sets are compact. Thus theorem 3 follows from theorem 2 applied to $(X, d^{'})$ . $□$

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Commentary