The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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

Proof. Since the collection of all quasi-compact opens forms a basis for the topology on $X$ and $X$ is sober, it is clear that $X$ is Hausdorff 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 $X$. 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. $\square$


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