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In the category $\text{Top}$ of topological spaces, start with the subcategory $\text{Disc}$ of spaces equipped with the discrete topology (which is equivalent to $\text{Set}$). Then take its closure under both limits and colimits (more precisely, consider the smallest subcategory of $\text{Top}$ containing the discrete spaces which is closed under both limits and colimits).

What topological spaces do you get? What category do you get?

Here is a guess: first, $\text{Disc}$ itself is closed under colimits. If you close it under limits only I think you get the prodiscrete spaces, or abstractly the pro-category $\text{Pro}(\text{Set})$. Then I think if you further close under colimits you get something like the "ind-prodiscrete spaces," or abstractly the ind-pro-category $\text{Ind}(\text{Pro}(\text{Set}))$. And maybe this thing is now closed under limits and colimits? I have no idea, really.

Whatever the abstract category is, I am separately interested in knowing what topological spaces can be obtained this way. For example, I think we can obtain $\mathbb{R}$ with the Euclidean topology! This is because we can obtain the Cantor set (it's the infinite product $\{ 0, 1 \}^{\mathbb{N}}$) and every compact metric space is a quotient of the Cantor set, so we can obtain the closed intervals $[-n, n]$, nest them in each other, and take the colimit. More generally I think we can get every ind-(compact Hausdorff) space via a similar argument. This suggests the category is more complicated than $\text{Ind}(\text{Pro}(\text{Set}))$.


Motivation (feel free to skip): This is partly an attempt to understand the following phenomenon. We know that the axiom of choice allows us to do strange things like write down strange ideals of the product $\prod_i F_i$ of countably many fields, whose quotients are strange rings like the ultraproducts. This is sometimes useful (for example it's useful to know that the ultraproduct of the fields $\mathbb{F}_p$ has characteristic zero, which lets us transfer certain statements over the fields $\mathbb{F}_p$ to characteristic zero) but arguably "pathological," and interestingly it seems like there are two different ways to rule these out:

  1. Don't use the axiom of choice. In ZF I believe it might be consistent that the only homomorphisms from a product $\prod_i F_i$ of fields into another field factor through the projections, although don't quote me on that.

  2. Equip $\prod_i F_i$ with the product topology and only quotient by closed ideals. This requires that we leave the setting of abstract rings and work with some flavor of topological rings, and this particular case we can handle by working with prodiscrete rings, so $\text{Pro}(\text{Set})$ suffices.

This is an example of a general phenomenon where in the absence of the axiom of choice it seems like various mathematical structures carry "canonical topologies" such that the only maps we can write down involving them are the continuous ones. One precise example of this phenomenon is that there's a model of ZF in which every homomorphism between separable completely metrizable topological groups is automatically continuous, so for example every homomorphism $\mathbb{R} \to \mathbb{R}$ of abelian groups is automatically continuous and there aren't any pathological solutions to the Cauchy functional equation. This might cover the above case if the $F_i$ are countable although I haven't checked this.

Another class of examples of important mathematical objects that seem to carry "canonical topologies" are infinite Galois groups, which are canonically profinite, so live in $\text{Pro}(\text{FinSet})$. The topology is crucial to stating the Galois correspondence in the infinite case, and furthermore it seems like every "natural" map you care to write down involving a Galois group is automatically continuous (again I think the above result might imply a version of this as a special case but I haven't checked).

These examples (and others) suggest to me that somehow algebra really does not want to take place entirely in $\text{Set}$; at least in the presence of the axiom of choice $\text{Set}$ allows too many irrelevant "garbage" maps (e.g. that spoil the infinite Galois correspondence). Instead somehow algebra wants to take place in a richer category, closer to but possibly not exactly $\text{Top}$; for example the category of condensed sets might be a better candidate. Or rather, it's as if instead of working in $\text{Set}$ we were really working with the subcategory of discrete spaces in $\text{Top}$ (or condensed sets, etc.) and various constructions (notably infinite limits) naturally leave this subcategory and land us in a richer category.

So, this question is an attempt to probe what happens if we literally do that, start with the discrete spaces and apply various constructions. I would also be very interested to know the answer to the same question for condensed sets, especially if it differs from $\text{Top}$.

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    $\begingroup$ Just to confirm: Every maximal ideal in a product of fields corresponds to an ultrafilter on the index set. There are models of ZF in which there are no non-principal ultrafilters on $\omega$, so in such a model, every homomorphism from a countable product of fields to another field factors through one of the projections (i.e., its kernel corresponds to a principal ultrafilter). $\endgroup$ Commented Jun 22 at 4:23
  • $\begingroup$ @Alex: nice, thank you! $\endgroup$ Commented Jun 22 at 4:25
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    $\begingroup$ It's not necessary to even start with discrete spaces--you could just say you are asking for the subcategory of Top generated by the empty subcategory under limits and colimits! $\endgroup$ Commented Jun 22 at 5:30
  • $\begingroup$ @Eric: right, that's a funny observation. So first we start with the empty subcategory and close under limits, and we get the one-point space $1$. Then we close under colimits and get $\text{Disc}$. So the process can be continued backwards two steps. $\endgroup$ Commented Jun 22 at 5:56

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You get all of Top. First, let me note that your subcategory is closed under taking subspaces. Indeed, if $X$ is in your category and $A\subseteq X$, the pushout $X\coprod_A X$ is also in your category (as the colimit of the diagram with two copies of $X$ and a singleton for each point of $A$ mapping into each of the copies). But now the equalizer of the two inclusions $X\to X\coprod_A X$ is just $A$.

So, since your subcategory contains $[0,1]$, it contains all completely regular spaces, since every completely regular space embeds in a power of $[0,1]$. Every topological space is a quotient of a completely regular space (see here for instance), and it follows that every topological space is in your subcategory.

(In fact, going through $[0,1]$ is unnecessary in this argument, since actually the argument I linked shows that any topological space is a quotient of a space whose topology is generated by clopen sets, so that it embeds in a power of $\{0,1\}$.)

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  • $\begingroup$ Fantastic! Let me just fill in a detail. We need to know that the subcategory is closed under quotients in the sense that if $X$ is in the subcategory and $Y$ is any quotient of it then $Y$ is in the subcategory. A priori this isn't clear because to take quotients categorically we need to consider the coequalizer of the equivalence relation regarded as a subspace of $X \times X$ and a priori this subspace isn't in the category even if $X$ is. But your observation about subspaces fixes this. $\endgroup$ Commented Jun 22 at 5:59
  • $\begingroup$ This seems to imply that the sort of ind-pro tower where we take the ind-category, then the pro-category, etc. stops in $\text{Top}$ after a finite number of steps? $\endgroup$ Commented Jun 22 at 6:00
  • $\begingroup$ Ah wait, it's simpler than that, I guess we can just use your trick of giving the equivalence relation the discrete topology? $\endgroup$ Commented Jun 22 at 6:05

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