Lemma 5.23.2. Let $X$ be a spectral space. The constructible topology is Hausdorff, totally disconnected, and quasi-compact.

Proof. Let $x, y \in X$ with $x \not= y$. Since $X$ is sober, there is an open subset $U$ containing exactly one of the two points $x, y$. Say $x \in U$. We may replace $U$ by a quasi-compact open neighbourhood of $x$ contained in $U$. Then $U$ and $U^ c$ are open and closed in the constructible topology. Hence $X$ is Hausdorff in the constructible topology because $x \in U$ and $y \in U^ c$ are disjoint opens in the constructible topology. The existence of $U$ also implies $x$ and $y$ are in distinct connected components in the constructible topology, whence $X$ is totally disconnected in the constructible topology.

Let $\mathcal{B}$ be the collection of subsets $B \subset X$ with $B$ either quasi-compact open or closed with quasi-compact complement. If $B \in \mathcal{B}$ then $B^ c \in \mathcal{B}$. It suffices to show every covering $X = \bigcup _{i \in I} B_ i$ with $B_ i \in \mathcal{B}$ has a finite refinement, see Lemma 5.12.15. Taking complements we see that we have to show that any family $\{ B_ i\} _{i \in I}$ of elements of $\mathcal{B}$ such that $B_{i_1} \cap \ldots \cap B_{i_ n} \not= \emptyset$ for all $n$ and all $i_1, \ldots , i_ n \in I$ has a common point of intersection. We may and do assume $B_ i \not= B_{i'}$ for $i \not= i'$.

To get a contradiction assume $\{ B_ i\} _{i \in I}$ is a family of elements of $\mathcal{B}$ having the finite intersection property but empty intersection. An application of Zorn's lemma shows that we may assume our family is maximal (details omitted). Let $I' \subset I$ be those indices such that $B_ i$ is closed and set $Z = \bigcap _{i \in I'} B_ i$. This is a closed subset of $X$ which is nonempty by Lemma 5.12.6. If $Z$ is reducible, then we can write $Z = Z' \cup Z''$ as a union of two closed subsets, neither equal to $Z$. This means in particular that we can find a quasi-compact open $U' \subset X$ meeting $Z'$ but not $Z''$. Similarly, we can find a quasi-compact open $U'' \subset X$ meeting $Z''$ but not $Z'$. Set $B' = X \setminus U'$ and $B'' = X \setminus U''$. Note that $Z'' \subset B'$ and $Z' \subset B''$. If there exist a finite number of indices $i_1, \ldots , i_ n \in I$ such that $B' \cap B_{i_1} \cap \ldots \cap B_{i_ n} = \emptyset$ as well as a finite number of indices $j_1, \ldots , j_ m \in I$ such that $B'' \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset$ then we find that $Z \cap B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset$. However, the set $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m}$ is quasi-compact hence we would find a finite number of indices $i'_1, \ldots , i'_ l \in I'$ with $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} \cap B_{i'_1} \cap \ldots \cap B_{i'_ l} = \emptyset$, a contradiction. Thus we see that we may add either $B'$ or $B''$ to the given family contradicting maximality. We conclude that $Z$ is irreducible. However, this leads to a contradiction as well, as now every nonempty (by the same argument as above) open $Z \cap B_ i$ for $i \in I \setminus I'$ contains the unique generic point of $Z$. This contradiction proves the lemma. $\square$

Comment #4224 by Laurent Moret-Bailly on

I suggest to add the property that the constructible topology is totally disconnected. (Proof: for $x$ and $y$ in $X$ and (say) $y$ not a specialization of $x$, there is a quasicompact open $U$ containing $y$ but not $x$. Now observe that $U$ is clopen in the constructible topology.) With the other properties this shows that the constructible topology is spectral.

Comment #4226 by Laurent Moret-Bailly on

Suggestion for a cleaner proof of quasi-compactness: let $\mathscr{P}$ be the set of finite partitions of $X$ into quasicompact, locally closed subspaces. For $P\in\mathscr{P}$, put $X_P:=\coprod_{Z\in P}Z$. Then observe that the constructible topology is $\varprojlim_{P\in\mathscr{P}}X_P$. This construction carries over nicely to schemes, showing that for a scheme $X$, $X^{\mathrm{cons}}$ has a natural scheme structure, giving a right adjoint to the inclusion (absolutely flat schemes)$\to$(schemes). Explicitly, let $X=\mathrm{Spec}(R)$. For $f\in R$, put $R\langle f\rangle=R_f\times R/f$. Then $X^{\mathrm{cons}}=\mathrm{Spec}(R^{\mathrm{cons}})$ where $R^{\mathrm{cons}}=\bigotimes_{f\in R}R\langle f\rangle$. The construction localizes well, giving rise to an affine morphism $X^{\mathrm{cons}}\to X$ for general $X$. It loooks simpler than the constructions given by Olivier and Hochster.

Comment #4406 by on

OK, I added the totally disconnected thing. See here.

Mostly it seems that one uses the quasi-compactness in proofs later on.

I feel the argument proving quasi-compactness is totally standard and straightforward. To do the argument you suggest for this, one needs to show that the pieces of a stratification P of the type you suggest are constructible, that one gets a cofiltered limit which involves some (easy) combinatorial arguments and then one needs to show that $X_{cons} = \lim X_P$, then one needs to apply Tychonov's thm, etc, etc. All of these things are easy, but I'm guessing it ends up being longer.

The construction for affine schemes is well known and is in the chapter on the pro-etale topology.

Comment #4858 by Kazuki Masugi on

There is no proof for the case $Z=\emptyset$.

Comment #5055 by Laurent Moret-Bailly on

Since the class of constructible sets is closed under finite intersections, the open (resp. closed) sets of the constructible topology are simply the unions (resp. intersections) of constructible sets. This provides a useful description of that topology, which could be mentioned explicitly.

Comment #5146 by on

@#4858: Thanks for pointing this out! @#5055: I added a sentence explaining this. For changes see here.

Comment #5926 by Yang Pei on

In the proof that the constructible topology on spectral space is quasi-compact, I think itâ€˜s worth pointing out that if all $B_i$ in $\mathcal{B}$ is quasi-compact open, in other words, $I'$ is empety, then $Z$ should by replaced by $X$ and $X$ is irreducible by the maximality of $\mathcal{B}$.

Comment #5927 by Yang Pei on

Sorry, I am wrong, $I'$ is not empty since at least $X$ is a complement of a quasi-compact open, which is closed.

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