$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Unif{Unif}\DeclareMathOperator\CHaus{CHaus}\DeclareMathOperator\Set{Set}\DeclareMathOperator\op{op}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Cond{Cond}\DeclareMathOperator\Top{Top}$In Barwick–Haine - Pyknotic objects, I. Basic notions Example 2.1.10, they showed that the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$ is a pyknotic set, i.e., a sheaf on the site $\CHaus$ of compact Hausdorff spaces equipped with the coherent topology (Lurie - Ultracategories).

I did not check whether the sheaf above is accessible (update: it is just the underlying topological space, see the **Update** part). However, I am slightly skeptical about this approach. It seems to me that the "correct" realization of a uniform space should be a condensed set which records its underlying topological space, along with an extra structure which records the uniform structure.

**Question:** I wonder how much is known about any kind of realization of uniform spaces as condensed sets.

I suppose that this extra structure would be described by a certain kind of enriched groupoid. Indeed, the uniform structure on a topological space could be understood as a groupoid enriched in filters. See nLab page for a description of this sort.

This is motivated by an attempt to eliminate the restriction that the adjective "solid" only applies to condensed *abelian* groups.

Let $M$ be a topological abelian group. I was about to understand what it means for $M$ that the condensed abelian group $\underline M$ is solid. Following Lecture II of Scholze - Lectures on Analytic Geometry, for any sequence $(m_n)_{n\in\mathbb N}\in M^{\mathbb N}$ convergent to $0$, we associate a (continuous) map from the profinite set $S\mathrel{:=}\mathbb N\cup\{\infty\}$ to $M$ which maps $n$ to $m_n$ and $\infty$ to $0$, or equivalently, a map $S\to\underline M$ of condensed sets by, say, Yoneda's lemma.

Suppose that $\underline M$ is solid, then this map extends uniquely to a map $\mathbb Z[S]^\blacksquare\to\underline M$ of condensed abelian groups. If I am not mistaken, $\mathbb Z[S]^\blacksquare\to M$ further factors through $\underline{\mathbb Z[[t]]}$, the condensed abelian group associated to the topological abelian group $\mathbb Z[[t]]$ with $(t)$-adic topology, and by full faithfulness, we get a factorization $S\to\mathbb Z[[t]]\to M$ where the second map is additive.

Consequently, for every sequence $(a_n)_{n\in\mathbb N}\in{\mathbb Z}^{\mathbb N}$ of integers, the series $\sum_na_nt^n$ converges in $\mathbb Z[[t]]$, therefore the series $\sum_na_nm_n$ converges in $M$, which should imply, if I am not mistaken, that the uniform structure on $M$ is non-archimedean and complete, at least when $M$ is first countable (by the way, I don't understand why it is claimed that “it is not directly as any kind of limit of finite sums”).

So the non-archimedean nature is rooted in the formalism. I suppose that a natural approach is to generalize the uniform structure to condensed sets, and to generalize the classical Cauchy-completeness. I don't know whether it is convinced that this does not work. The current presentation separates non-archimedean and $\mathbb R$-case, which covers neither non-abelian groups nor the general completeness of topological abelian groups.

**Update:** When discussing the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$, I missed the simple fact that $\Hom_{\Unif}(-,X)=\Hom_{\Top}(-,X)\mathrel{=:}\underline X$ by the Heine–Cantor theorem. In other words, when the uniform space in question is $T_0$ (therefore $T_1$), $\underline X$ is thus a condensed set.

**Update2:** It seems to me that I did not phrase the question unambiguously. In fact, I wanted to ask for any realization of unform spaces as condensed sets *with some extra data*, which should be recorded in the data of a condensed abelian group when it comes from a topological abelian group. At the time that I posed the question, I did not realize that Barwick–Haine's approach only records the data about the underlying space, but just doubted that it is a "correct" approach to record all data.

**Update3:** We say that a uniform space is *non-archimedean* if the uniformity is generated by equivalence relations.