Lemma 5.22.4 (08ZZ)—The Stacks project

Lemma 5.22.4. Let $X$ be a profinite space. Every open covering of $X$ has a refinement by a finite covering $X = \coprod U_ i$ with $U_ i$ open and closed.

Proof. Write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a limit of an inverse system of finite discrete spaces over a directed set $I$ (Lemma 5.22.2). Denote $f_ i : X \to X_ i$ the projection. For every point $x = (x_ i) \in X$ a fundamental system of open neighbourhoods is the collection $f_ i^{-1}(\{ x_ i\} )$. Thus, as $X$ is quasi-compact, we may assume we have an open covering

\[ X = f_{i_1}^{-1}(\{ x_{i_1}\} ) \cup \ldots \cup f_{i_ n}^{-1}(\{ x_{i_ n}\} ) \]

Choose $i \in I$ with $i \geq i_ j$ for $j = 1, \ldots , n$ (this is possible as $I$ is a directed set). Then we see that the covering

\[ X = \coprod \nolimits _{t \in X_ i} f_ i^{-1}(\{ t\} ) \]

refines the given covering and is of the desired form. $\square$

Comments (4)

Comment #7038 by Amnon Yekutieli on February 10, 2022 at 15:35

@Johan: There seems to be a problem in the proof of Lemma 08ZZ. I think I can pinpoint it: there is no explanation why the covering produced refines the original covering. Moreover, it probably does not. This is because the open sets f_i^{-1}({x_i}) do not form a basis of open neighborhoods of X, only a sub-basis; it is necessary to take finite intersections to obtain a basis. A possible fix was sent to you by email.

Comment #7039 by Johan on February 10, 2022 at 15:44

So, I probably should think more, but I think the argument is OK as written. If $x = (x_i)$ is a point of $X$ and if $U \subset X$ is an open containing $x$ , then for some $i$ large enough we see that $f^{-1}(\{x_i\})$ is contained in $U$ . So any open covering of $X$ can be refined by a (possibly infinite) open covering whose members are of this form. Since $X$ is quasi-compact, you can assume the covering is finite. OK?

Comment #7043 by Amnon Yekutieli on February 11, 2022 at 12:27

@Johan: My mistake. I forgot to take into account that the X_i form an inverse system (I only treated the product space). Still I think it would be better to add a few more words.

I typed a proof based on your proof but with more details (link below). The part in yellow highlight should somehow be mentioned in your proof, I think. \ref{https://drive.google.com/file/d/1duVZeEs7smLoVlZX0YFfqFonXLBxzGgG/view?usp=sharing}

Comment #7045 by Johan on February 11, 2022 at 12:41

OK, Amnon, sounds good. (To everybody else, the Amnon kindly sent me a pdf containing the part in yellow he mentioned; I apologize to everybody for the difficulty of getting the comments to look good! Links can be added but you have to use the [link text](url) construction!)

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