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$\DeclareMathOperator\colim{colim}$This is inspired by Clausen's answer.

Question: Recall that $\mathbb Z_p$ is endowed with the $p$-adic topology. Consider the countable sum $M:=\bigoplus_{n=0}^\infty\mathbb Z_p$ as a topological abelian group. One way to describe the topology is by the filtered colimit (or "inductive/direct limit topology" in the classical literature) $\colim_{N\in\mathbb N}\bigoplus_{n=0}^N\mathbb Z_p$.

  1. As a topological space, is $M$ first-countable? (I guess not)
  2. As a topological group (or more generally, a uniform space), is $M$ Cauchy-complete (i.e., every Cauchy filter converges)?

First, let's remark that $M$ is Hausdorff: for any two elements $x,y\in M$, there exists an $N\in\mathbb N$ such that $x,y$ belong to $\bigoplus_{n=0}^N\mathbb Z_p$. The result then follows from the fact that $M$ is isomorphic to $\left(\bigoplus_{n=0}^N\mathbb Z_p\right)\oplus\left(\bigoplus_{n=N+1}^\infty\mathbb Z_p\right)$. We also note that $M$ is compactly generated by definition.

We usually take the $p$-adic completion $\left(\bigoplus_{n=0}^\infty\mathbb Z\right)_p^\wedge$ in algebra, which leads to the impression that $M$ is not complete. However, after some thoughts, I find that the situation is not that simple: the opens in $M$ might be very complicated.

To see this, let $B$ denote the countable product $B:=\left(\prod_{n=0}^\infty\right)'\mathbb Z_p$ endowed with the box topology instead of the product topology. It follows, say from universal property of filtered colimit, that there exists a continuous map $M\to B$ which maps $\sum_n j_n(x_n)$ to $(x_0,x_1,\ldots)$ where $\sum_n j_n(x_n)$ is a finite sum and $j_n\colon\mathbb Z_p\to M$ is the canonical map. $B$ is not first countable, which leads me to the suspicion that neither is $M$.

Furthermore, from experience in algebra, one might mistakenly think that the series $\sum_{n=0}^\infty j_n(p^n)$ "should be" Cauchy therefore $M$ is not complete. The problem is that $j_n(p^n)$ does not converge to $0$. Indeed, $\bigoplus_{n=0}^\infty p^{n+1}\mathbb Z_p\subseteq M$ is an open neighborhood of $0$ in $M$ (which follows either from the continuity of $M\to B$, or a direct verification) to which none of $j_n(p^n)$ belongs.


Relation to condensed mathematics: it follows from Barwick-Haine Lemma 2.1.7 and Scholze's notes Lemma 1.3 that the countable direct sum $M$ coincides with the countable direct sum taken in the category of condensed abelian groups. Since $\mathbb Z_p$ is solid, so is the direct sum. It then follows from my argument that $M$ is sequentially Cauchy-complete, i.e. every Cauchy sequence in $M$ converge.


Summary: As Scholze indicated in the comment at his answer, one could replace the countable sum by arbitrary sum. One need to show that $\{U_f\}$ in that answer is indeed a basis of the topology, which follows from Tube lemma (note that $\mathbb Z_p$ is profinite therefore $p^m\mathbb Z_p$ is CHaus) and Zorn's lemma.

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The sum $M=\bigoplus_{\mathbb N} \mathbb Z_p$ is not first-countable, but it is Cauchy complete. More precisely, $M$ maps isomorphically to $\varprojlim_{U\subset M} M/U$ where $U$ runs over open subgroups (so $M/U$ is discrete). This was observed in Example 7.1.7 here (and surely elsewhere before).

Let's describe a cofinal system of open subgroups of $M$. These are given by $U=U_f=\bigoplus_{\mathbb N} p^{f(n)} \mathbb Z_p$ for some function $f: \mathbb N\to \mathbb N$, arbitrarily fast increasing. As for any countable collection of $f$'s, one can find a function that is eventually larger than each of them, there is no countable neighborhood basis of $0$. On the other hand, the map $M\to \varprojlim_f M/U_f$ is an isomorphism. Indeed, one easily sees that the inverse limit injects into $\prod_{\mathbb N} \mathbb Z_p$, and for any sequence of elements of $\mathbb Z_p$ that is not eventually $0$, one can find some $f$ such that the image in $M/U_f$ is still not zero.

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  • $\begingroup$ I think the same argument applies to any (possibly uncountable) direct sum of $\mathbb Z_p$'s as well, and even in that case the topological and condensed versions match; more precisely, any continuous map $S\to \bigoplus \mathbb Z_p$ factors over a finite direct sum. This follows from Proposition A.15 here. Still, these groups feel much more natural from a condensed point of view -- compare a countable direct sum with a weird uncountable limit. $\endgroup$ Mar 24, 2021 at 13:49
  • $\begingroup$ Thanks for this, Peter and Z. M: I'd naively guessed it wasn't Cauchy complete! I guess solidness and Cauchy completeness are more closely related than I thought. $\endgroup$ Mar 24, 2021 at 15:13
  • $\begingroup$ @PeterScholze: Is this the same as taking the restricted product topology with the subspaces $\{0\}\subset\mathbb{Z}_p$? $\endgroup$ Mar 24, 2021 at 19:03
  • $\begingroup$ Hmm, yes, I guess it is? But to me, it seems like a strange way to think about it. $\endgroup$ Mar 24, 2021 at 19:32
  • $\begingroup$ Indeed. But it also seems to me a way to put under the same roof the adeles (I was wondering where did someone show they are not metrizable as in this MO post) and $\mathscr{D}(\Omega)$ the space of test functions, or more prosaically the space $\mathbb{R}[X]$ of polynomials in one variable. $\endgroup$ Mar 24, 2021 at 20:36

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