# Can locally constant real functions on a space be made into continuous functions (on a different space)?

[This question came up while idly thinking about this other one, but it is not directly related.]

Definitions: If $$X$$ is a topological space, let $$C(X)$$ stand for the $$\mathbb{R}$$-algebra of continuous real-valued functions $$X\to\mathbb{R}$$ where $$\mathbb{R}$$ has its usual (Euclidean) topology, and let $$D(X)$$ stand for the $$\mathbb{R}$$-algebra of locally constant functions $$X\to\mathbb{R}$$, or equivalently, continuous functions $$X\to\mathbb{R}_{\mathrm{disc}}$$ where $$\mathbb{R}_{\mathrm{disc}}$$ stands for $$\mathbb{R}$$ endowed with the discrete topology. Both $$C$$ and $$D$$ can be seen as contravariant functors from the category of topological spaces to the category of $$\mathbb{R}$$-algebras (by taking a continuous map $$f\colon X\to Y$$ to the ring homomorphism defined by right-composition by $$f$$).

Motivations for the following questions: Quite a lot is known about $$C(X)$$ (and its subalgebra $$C^*(X)$$ of bounded functions), as witnessed, e.g., by the classic book by Gillman & Jerison, Rings of Continuous Functions (and also its “sequel” of sorts, by Fine, Gillman & Lambek, Rings of Quotients of Rings of Functions). For example, $$C(X)$$ uniquely determines $$X$$ up to isomorphism when $$X$$ is [completely regular and] realcompact, and when this is the case, $$X$$ can be recovered as the closure of the image of $$X$$ under the evaluation map $$X \to \mathbb{R}^{C(X)}$$, or as the set of maximal ideals of $$C(X)$$ whose quotient ring is $$\mathbb{R}$$ endowed with the Zariski topology. However, that I know of, there is no satisfactory purely algebraic characterization of $$\mathbb{R}$$-algebras of the form $$C(X)$$.

Now I realized to my horror that I had no idea about the analogous questions for $$D(X)$$ (and its subalgebra $$D^*(X) := D(X) \cap C^*(X)$$). I could ask a million questions, the gist of which would be “where can I learn about the same things concerning $$D(X)$$ that Gillman & Jerison teaches us about $$C(X)$$?”, so if somebody knows the answer to that, please do tell; but since MathOverflow requires that I ask something more specific than “every question answered in Gillman & Jerison”, let me ask something slightly different, namely whether we can reduce the study of $$D(X)$$ to that of the well understood $$C(X)$$. Namely:

Questions: Given a topological space $$X$$, does there always exist a topological space $$\nabla X$$ such that $$C(\nabla X) = D(X)$$? Better: can we find a functor $$\nabla$$ (from topological spaces to topological spaces) such that $$C\circ\nabla = D$$ and a natural transformation $$\eta \colon 1_{\mathbf{Top}} \to \nabla$$ such that $$C(\eta_X)\colon C(\nabla X) = D(X) \to C(X)$$ is the inclusion, and perhaps, say, that $$\eta$$ is an isomorphism on discrete spaces? Even better: is $$\nabla$$ left-adjoint to the inclusion functor of the full subcategory of topological spaces $$X$$ for which $$D(X)=C(X)$$ (the existence of this left adjoint would answer all questions positively)? [edit: see below]

PS: I just remembered that spaces such that $$D(X)=C(X)$$ are known as “P-spaces”, but this doesn't really help. It does, however, allow me to restate the strongest version of my question as: “the the full subcategory of P-spaces reflective?”. Also, maybe I should be formulating my questions in the language of locales rather than topological spaces?

Edit: Simon Pepin Lehalleur has just pointed out to me that an arbitrary product of P-spaces can fail to be a P-space (Gillman & Jerison, exercise 4K(6)), so we can't hope for them to form a reflective subcategory. This doesn't invalidate weaker forms of the question, however.

Remark: Even in the case of $$X=\mathbb{Q}$$ (with the usual (=order) topology), I don't know whether there exists a space $$\nabla\mathbb{Q}$$ such that $$C(\nabla\mathbb{Q}) = D(\mathbb{Q})$$.

• It is more natural to look at compactly supported functions than bounded or arbitrary ones I think. If K is a field then the rings of compactly supported locally constant maps to K are precisely the commutative K-algebras spanned by idempotents by a result of Keimmel. Here the space is the maximal ideal space but can also be viewed as the Stone space of the boolean algebra of idempotents. It is locally compact and totally disconnected Hausdorff. I don't require unital. Jan 23, 2021 at 21:19
• @LSpice Maybe there's some confusion as to the meaning of “locally constant” (maybe you're thinking step functions?), but $D(X) \subseteq C(X)$, and, indeed, $D(X)$ (and *a fortriori $D^*(X)$) consists only of constants for $X$ connected. Jan 23, 2021 at 22:20
• Sorry, yes. I don't know what I was thinking (and appreciate your charitable attempt to sensible-ify it), and will delete my silly comment. Jan 23, 2021 at 22:22
• What do you denote by equality in your last remark? An $\mathbf{R}$-algebra isomorphism? preserving positives? with more "natural" properties?
– YCor
Jan 24, 2021 at 7:50
• @SimonPepinLehalleur Sorry! Fixed. Jan 24, 2021 at 20:31

$$D(\mathbb{Q})$$ is not isomorphic to $$C(X)$$ (as an $$\mathbb{R}$$-algebra) for any topological space $$X$$. For both $$D(X)$$ and $$C(X)$$, you can recover the (extended) uniform norm from the $$\mathbb{R}$$-algebraic structure with $$\left\| x \right\| = \sup\{|\lambda| : x - \lambda 1\text{ is not invertible}\},$$ where $$1$$ is the multiplicative identity of the algebra and $$\lambda$$ ranges over $$\mathbb{R}$$.
$$D(\mathbb{Q})$$ is not metrically complete under the (extended) metric induced by the uniform norm, but $$C(X)$$ always is, so $$D(\mathbb{Q})$$ can't be isomorphic to any $$C(X)$$ as an $$\mathbb{R}$$-algebra.
• Great! So it seems I will now have to ask every question about $D(X)$ separately, since they can't be reduced to $C(X)$. 😕 Jan 23, 2021 at 22:17