$\DeclareMathOperator\colim{colim}$This is inspired by [Clausen's answer](https://mathoverflow.net/a/387158/).


**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)?




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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](https://en.wikipedia.org/wiki/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.




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Relation to condensed mathematics: it follows from [Barwick-Haine](https://arxiv.org/pdf/1904.09966.pdf) Lemma 2.1.7 and [Scholze's notes](https://www.math.uni-bonn.de/people/scholze/Analytic.pdf) 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](https://mathoverflow.net/q/387055/) that $M$ is sequentially Cauchy-complete, i.e. every Cauchy sequence in $M$ converge.




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**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](https://en.wikipedia.org/wiki/Tube_lemma) (note that $\mathbb Z\_p$ is profinite therefore $p^m\mathbb Z\_p$ is CHaus) and Zorn's lemma.



Countable sum $\bigoplus_{n=0}^\infty\mathbb Z_p$ as a topological group
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](https://www.math.uni-bonn.de/people/scholze/proetale.pdf) (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.


