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Peter Scholze
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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 algebraic geometryAnalytic 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.

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 algebraic 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.

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.

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LSpice
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$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Unif{Unif}\DeclareMathOperator\CHaus{CHaus}\DeclareMathOperator\Set{Set}\DeclareMathOperator\op{op}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Cond{Cond}\DeclareMathOperator\Top{Top}$In BarwickBarwick–Haine -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 (coherent topologyLurie - Ultracategories).

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 Lecture IIScholze - Lectures on algebraic 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:=\mathbb N\cup\{\infty\}$$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.

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“it is not directly as any kind of limit of finite sums”).

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)=:\underline X$$\Hom_{\Unif}(-,X)=\Hom_{\Top}(-,X)\mathrel{=:}\underline X$ by the Heine-CantorHeine–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'sBarwick–Haine's approach only records the data about the underlying space, but just doubted that it is a "correct" approach to record all data.

$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Unif{Unif}\DeclareMathOperator\CHaus{CHaus}\DeclareMathOperator\Set{Set}\DeclareMathOperator\op{op}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Cond{Cond}\DeclareMathOperator\Top{Top}$In Barwick-Haine 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.

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, 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 $S:=\mathbb N\cup\{\infty\}$ 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.

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).

Update: When discussing the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$, I missed the simple fact that $\Hom_{\Unif}(-,X)=\Hom_{\Top}(-,X)=:\underline X$ 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.

$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Unif{Unif}\DeclareMathOperator\CHaus{CHaus}\DeclareMathOperator\Set{Set}\DeclareMathOperator\op{op}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Cond{Cond}\DeclareMathOperator\Top{Top}$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).

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 algebraic 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.

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”).

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.

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

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

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