I was planning to continue the subject of exponentiable locales, but that grew out of hand. Instead, let me touch a lighter subject today. If you thought that my posts were more and more complex each time, that should be a relief to you. If you wish to complain that I am not giving you enough thought for food, or that I am becoming lazy, please consider that I posted twice last month, and that the post on the seminar on continuity in semilattices points to 98 papers for you to peruse, no less! 😀
Separate and joint continuity
The question we will address today is in the following setting. We consider a function f : X × Y → Z, where X, Y, Z are topological spaces, and we know that f is separately continuous; we would like to prove that f is jointly continuous. When is that true?
You have probably been warned that there are functions in two arguments in nature that are separately continuous but not jointly continuous, and Exercise 4.5.11 in the book is a classical counterexample. However, there are cases where separate continuity implies (and therefore, is equivalent to) joint continuity.
The Ershov observation
I first learned of the following result, due to Yuri Ershov [1, Proposition 2] from Klaus Keimel, who used it to show that if the scalar multiplication on a cone C (whatever that is; this is not important), which happens to be a map from R+ × C to C, is separately continuous, then it is automatically jointly continuous [2]. Here R+ has to have the Scott topology of its usual ordering ≤.
For some time, I called the following the Ershov observation. A c-space is a space X in which for every point x, for every open neighborhood U of x, there is a point y in U such that x is in the interior of the upward closure ↑y of y; the sober c-spaces are exactly the continuous dcpos in their Scott topology. This applies in the case of scalar multiplication cited above because R+, with its usual Scott topology (not its metric topology) is a c-space; in fact, a continuous poset.
Proposition (Ershov). Let X be a c-space, and Y, Z be arbitrary topological spaces. A function f : X × Y → Z is separately continuous if and only if it is jointly continuous.
Let me give a proof of this fact. This is not hard. However, please be warned that we will give another theorem below which subsumes this one.
Proof. Let us assume that f is separately continuous. Let W be an open subset of Z, and (x, y) ∈ f −1(W). Since f is separately continuous, f (_, y) is continuous, and therefore f (_, y)−1(W) is an open neighborhood of x. We use the fact that X is a c-space and we obtain a point x’ ∈ f (_, y)−1(W) such that x ∈ int(↑x’).
Let us use the fact that f is separately continuous a second time, but this time on the other argument: f (x’, _) is continuous; so f (x’, _)−1(W) is open. We had built x‘ so that x’ ∈ f (_, y)−1(W), so f (x’, y) is in W, and hence y is in f (x’, _)−1(W).
Then int(↑x’) × V is an open neighborhood of (x, y) in the product topology, and it remains to show that it is included in f−1(W). For every pair (x”, y”) ∈ int(↑x’) × V, f (x”, y”) is larger than or equal to f (x’, y”) since x’ ≤ x” and since every continuous map is monotonic. By construction, f (x’, y”) is in W, and therefore f (x”, y”) is also in W, since W is upwards-closed. ☐
The Banaschewski-Lawson observation, part I
A few years later, I realized that some people (Zhenchao Lyu, Xiaolin Xie, and Hui Kou) knew better than I did: Jimmie Lawson had proved a rather sweeping generalization of the Ershov observation… twelve years earlier [3, Theorem 2]. I have already mentioned this in my post on the ISDT’22 conference. It is time I gave the proof.
Oh, if you read the paragraph before Theorem 2 in [3], you will realize that Jimmie Lawson says that most of that Theorem is due to Bernard Banaschewski. And indeed, the following is part of Proposition 6 of [4], which predates [3] by five more years. But then, as J. Lawson notices, Banaschewski’s Corollary 2 to his Proposition 3 is incorrect, and is used at some point… it is always so complicated to trace back the history of a result!
Proposition (Banaschewski-Lawson, part I). Let X be a locally finitary compact space, and Y, Z be arbitrary topological spaces. A function f : X × Y → Z is separately continuous if and only if it is jointly continuous.
This is exactly the same statement as with Ershov’s observation, except that X is only assumed to be locally finitary compact. In other words, for every point x of X, for every open neighborhood U of x, there is a finitary compact set ↑E (where E, by definition, is finite) such that x ∈ int(↑E) ⊆ ↑E ⊆ U. This is a strong form of local compactness, where the interpolating compact set Q (i.e., such that x ∈ int(Q) ⊆ ↑Q ⊆ U) has to be the upward closure of finitely many points. C-spaces are an even stronger form of local compactness, where E is required to contain exactly one point.
Proof. The proof is exactly as with Ershov’s observation, changing only the details that must change.
Let us assume that f is separately continuous. Let W be an open subset of Z, and (x, y) ∈ f −1(W). Since f is separately continuous, f (_, y) is continuous, and therefore f (_, y)−1(W) is an open neighborhood of x. We use the fact that X is locally finitary compact and we obtain a finite subset A of f (_, y)−1(W) such that x ∈ int(↑A)—instead of just a single point x’, as in the proof of Ershov’s observation.
Let us use the fact that f is separately continuous a second time, but this time on the other argument. For every x’ ∈ A, f (x’, _) is continuous, so f (x’, _)−1(W) is open. By definition of A, x’ is in f (_, y)−1(W), so f (x’, y) is in W, and hence y is in f (x’, _)−1(W).
Let V be the intersection of all the open sets f (x’, _)−1(W), when x’ ranges over A. Since A is finite, V is open. Also, y is in V.
Then int(↑A) × V is an open neighborhood of (x, y) in the product topology, and it remains to show that it is included in f−1(W). For every pair (x”, y”) ∈ int(↑A) × V, x” is in particular in ↑A, so there is a point x’ in A such that x’≤x”. Since every continuous map is monotonic, it follows that f (x”, y”) is larger than or equal to f (x’, y”). By construction, y” is in V, hence in f (x’, _)−1(W), so f (x’, y”) is in W. Therefore f (x”, y”) is also in W, since W is upwards-closed. ☐
The Banaschewski-Lawson observation, part II
What makes the Banaschewski-Lawson observation more remarkable is that the condition is not only sufficient, it is necessary. The following is the equivalence between items (6) and (8) in J. Lawson’s Theorem 2 in [3].
Theorem (Banaschewski-Lawson). The class C of topological spaces X such that for all topological spaces Y and Z, every separately continuous function f : X × Y → Z is jointly continuous is exactly the class of locally finitary compact spaces.
Proof. Let us call a good space any space X with the above property; namely, such that for all topological spaces Y and Z, every separately continuous function f : X × Y → Z is jointly continuous. The previous proposition states that every locally finitary compact space is good.
Conversely, let X be any good space. For Z, we take Sierpiński space S, the dcpo {0 < 1} with its Scott topology. For Y, we consider OpX, namely the set of all open subsets of X, with the pointwise topology (not the Scott topology). The latter is defined as being generated by the subbasic open sets [x ∈] ≝ {U ∈ OX | x ∈ U}, where x ranges over the points of X. We have already seen that space when we studied the double powerspace constructions.
Finally, we let f : X × Y → Z be the characteristic function of the membership relation ∈. In other words, f(x, U) is equal to 1 if x ∈ U, to 0 otherwise. We verify that this is a separately continuous function. There is only one non-trivial open set in Z, which is {1}, hence it suffices to check the inverse images of {1} under f(x,_) and under f(_,U).
- For every x ∈ X, f(x,_)−1({1}) is the collection of open neighborhoods of x, namely [x ∈]. This is open in OpX by definition of the pointwise topology.
- For every U ∈ OX, f(_,U)−1({1}) is equal to U, which is of course open.
Hence f is separately continuous. Since X is good, f must be jointly continuous, by definition. We wish to show that X is locally finitary compact. We consider any point x of X, and any open neighborhood U of x. Hence f(x,U)=1, so (x,U) is inside the open set f−1({1}). By definition of the product topology, there is an open rectangle V × W in X × OpX such that x ∈ V, U ∈ W, and V × W ⊆ f−1({1}).
Since W is open in OpX, by definition of the pointwise topology, W contains a finite intersection ∩k=1n [xk ∈] of subbasic open sets, such U is in every [xk ∈]. Let A be the finite set {x1, …, xn}.
- Since U is in [xk ∈] for every k, xk is in U for every k; in other words, A is included in U, or equivalently, ↑A is included in U.
- The open rectangle V × ∩k=1n [xk ∈] is included in V × W, which is included in f−1({1}). In other words, for every point x’ in V, for every open subset V’ of X that belongs to ∩k=1n [xk ∈], f(x’,V)=1; equivalently, x’ is in V. The open subsets V’ of X that belong to ∩k=1n [xk ∈] are those such that xk ∈ V’ for every k, and are therefore exactly the open neighborhoods of A. Hence we have shown that for every point x’ in V, x’ is in every open neighborhood V’ of A. The intersection of those neighborhoods is the saturation ↑A of A. Therefore we have just shown that every point of V is is ↑A.
Since x is in V ⊆ ↑A, we conclude that x is in int(↑A).
Summing up, x ∈ int(↑A) ⊆ ↑A ⊆ U, completing the proof that X is locally finitary compact. ☐
The topology of separate continuity
I said that this was just the equivalence between items (6) and (8) of [3, Theorem 2]. That theorem really states the equivalence of no less than 12 conditions! Those extend the equivalence between 7 of those conditions given in [4, Proposition 6].
Let me mention just one of those extra equivalent conditions. Condition (7) in [3, Theorem 2], which is also condition (5) in [4, Proposition 6], is: “for every T0 space Y, X × Y = X ⊗ Y“. By definition, X ⊗ Y is the product of X and Y with the following topology: a subset W of X × Y is open in X ⊗ Y if and only if all the slices W|x ≝ {y ∈ Y | (x, y) ∈ W} and W|y ≝ {x ∈ X | (x, y) ∈ W} are open, for all x in X and y in Y. This topology is called the topology of separate continuity for the following reason.
Lemma. For all topological spaces X, Y and Z, a function f : X × Y → Z is separately continuous if and only if it is continuous from X ⊗ Y to Z.
Proof. For every open subset W of Z, and every point x in X, f–1(W)|x is equal to f(x,_)–1(W), and similarly, for every point y in Y, f(_,y)–1(W) = f–1(W)|y. Therefore, if f is separately continuous then f–1(W) is open in X ⊗ Y for every open subset W of Z, and conversely. ☐
As a consequence, given two topological spaces X and Y, the following statements are equivalent:
- for every topological space Z, every separately continuous function f : X × Y → Z is jointly continuous;
- X × Y = X ⊗ Y, where X × Y is given the usual product topology.
Indeed, if the former is true then by taking Z ≝ S, we obtain that the open subsets of X × Y are exactly the same as those of X ⊗ Y, since the first are those whose characteristic function is jointly continuous, and the second are those whose characteristic function is separately continuous, by the lemma we have just stated and proved. Conversely, if X × Y = X ⊗ Y, then every separately continuous function f : X × Y → Z is continuous from X ⊗ Y to Z, hence from X × Y (with the product topology), namely it is jointly continuous.
The following should now be clear.
Proposition. The topology of separate continuity (on X ⊗ Y) is finer than the product topology. The class of topological spaces X such that for every topological space Y, X × Y = X ⊗ Y, is exactly the class of locally finitary compact spaces.
We can replace “every topological space Y” by “every T0 space Y” in this statement, too, or by “for the topology space Y equal to OpX” (this is the only space we needed in the proof). The space OpX has several other properties; for example, it is sober (sorry, no proof of that for today), so we can also write “for every sober space Y” in the latter proposition, and so on, and so forth.
At any rate, let me finish this post here. I will let you discover the other 9 equivalent properties stated by J. Lawson in [3] by yourselves. Some are concerned with hypercontinuous lattices, some with the injectivity of certain function spaces, and so on. I said this post would be less complex than my recent posts, and I wish to keep this promise. The next post may be more complicated again!
- Yuri Leonidovich Ershov (Юрий Леонидович Ершов). The bounded-complete hull of an α-space. Theoretical Computer Science 175, 1997, pages 3-13.
- Klaus Keimel. Topological cones: functional analysis in a T0-setting. Semigroup forum 77(1), pages 109-142, 2008.
- Jimmie D. Lawson. T0-spaces and pointwise convergence. Topology and Its Applications, 21(1), 73–76, 1985.
- Bernard Banaschewski. Essential extensions of T0 spaces. General Topology and its Applications 7 (1977) 233–246. North-Holland Publishing Company.
— Jean Goubault-Larrecq (June 20th, 2023)