The Stacks Project


Tag 0A31

Chapter 5: Topology > Section 5.24: Limits of spectral spaces

Lemma 5.24.7. Let $W$ be a subset of a spectral space $X$. The following are equivalent:

  1. $W$ is an intersection of constructible sets and closed under generalizations,
  2. $W$ is quasi-compact and closed under generalizations,
  3. there exists a quasi-compact subset $E \subset X$ such that $W$ is the set of points specializing to $E$,
  4. $W$ is an intersection of quasi-compact open subsets,
  5. there exists a nonempty set $I$ and quasi-compact opens $U_i \subset X$, $i \in I$ such that $W = \bigcap U_i$ and for all $i, j \in I$ there exists a $k \in I$ with $U_k \subset U_i \cap U_j$.

In this case we have (a) $W$ is a spectral space, (b) $W = \mathop{\rm lim}\nolimits U_i$ as topological spaces, and (c) for any open $U$ containing $W$ there exists an $i$ with $U_i \subset U$.

Proof. Let $W \subset X$ satisfy (1). Then $W$ is closed in the constructible topology, hence quasi-compact in the constructible topology (by Lemmas 5.23.2 and 5.12.3), hence quasi-compact in the topology of $X$ (because opens in $X$ are open in the constructible topology). Thus (2) holds.

It is clear that (2) implies (3) by taking $E = W$.

Let $X$ be a spectral space and let $E \subset W$ be as in (3). Since every point of $W$ specializes to a point of $E$ we see that an open of $W$ which contains $E$ is equal to $W$. Hence since $E$ is quasi-compact, so is $W$. If $x \in X$, $x \not \in W$, then $Z = \overline{\{x\}}$ is disjoint from $W$. Since $W$ is quasi-compact we can find a quasi-compact open $U$ with $W \subset U$ and $U \cap Z = \emptyset$. We conclude that (4) holds.

If $W = \bigcap_{j \in J} U_j$ then setting $I$ equal to the set of finite subsets of $J$ and $U_i = U_{j_1} \cap \ldots \cap U_{j_r}$ for $i = \{j_1, \ldots, j_r\}$ shows that (4) implies (5). It is immediate that (5) implies (1).

Let $I$ and $U_i$ be as in (5). Since $W = \bigcap U_i$ we have $W = \mathop{\rm lim}\nolimits U_i$ by the universal property of limits. Then $W$ is a spectral space by Lemma 5.24.5. Let $U \subset X$ be an open neighbourhood of $W$. Then $E_i = U_i \cap (X \setminus U)$ is a family of constructible subsets of the spectral space $Z = X \setminus U$ with empty intersection. Using that the spectral topology on $Z$ is quasi-compact (Lemma 5.23.2) we conclude from Lemma 5.12.6 that $E_i = \emptyset$ for some $i$. $\square$

    The code snippet corresponding to this tag is a part of the file topology.tex and is located in lines 4514–4534 (see updates for more information).

    \begin{lemma}
    \label{lemma-make-spectral-space}
    Let $W$ be a subset of a spectral space $X$. The following are equivalent:
    \begin{enumerate}
    \item $W$ is an intersection of constructible sets and
    closed under generalizations,
    \item $W$ is quasi-compact and closed under generalizations,
    \item there exists a quasi-compact subset $E \subset X$ such that
    $W$ is the set of points specializing to $E$,
    \item $W$ is an intersection of quasi-compact open subsets,
    \item
    \label{item-intersection-quasi-compact-open}
    there exists a nonempty set $I$ and quasi-compact opens
    $U_i \subset X$, $i \in I$
    such that $W = \bigcap U_i$ and for all $i, j \in I$ there exists a
    $k \in I$ with $U_k \subset U_i \cap U_j$.
    \end{enumerate}
    In this case we have (a) $W$ is a spectral space, (b) $W = \lim U_i$
    as topological spaces, and (c) for any open $U$ containing $W$
    there exists an $i$ with $U_i \subset U$.
    \end{lemma}
    
    \begin{proof}
    Let $W \subset X$ satisfy (1). Then $W$ is closed in the constructible
    topology, hence quasi-compact in the constructible topology (by
    Lemmas \ref{lemma-constructible-hausdorff-quasi-compact} and
    \ref{lemma-closed-in-quasi-compact}), hence quasi-compact in the topology
    of $X$ (because opens in $X$ are open in the constructible topology). Thus
    (2) holds.
    
    \medskip\noindent
    It is clear that (2) implies (3) by taking $E = W$.
    
    \medskip\noindent
    Let $X$ be a spectral space and let $E \subset W$ be as in (3).
    Since every point of $W$ specializes to a point of $E$ we see that
    an open of $W$ which contains $E$ is equal to $W$. Hence since $E$
    is quasi-compact, so is $W$.
    If $x \in X$, $x \not \in W$, then $Z = \overline{\{x\}}$ is
    disjoint from $W$. Since $W$ is quasi-compact we can find a
    quasi-compact open $U$ with $W \subset U$ and $U \cap Z = \emptyset$.
    We conclude that (4) holds.
    
    \medskip\noindent
    If $W = \bigcap_{j \in J} U_j$ then setting $I$ equal to the set of
    finite subsets of $J$ and $U_i = U_{j_1} \cap \ldots \cap U_{j_r}$
    for $i = \{j_1, \ldots, j_r\}$ shows that (4) implies (5). It is immediate
    that (5) implies (1).
    
    \medskip\noindent
    Let $I$ and $U_i$ be as in (5).
    Since $W = \bigcap U_i$ we have $W = \lim U_i$ by the universal property
    of limits. Then $W$ is a spectral space by
    Lemma \ref{lemma-directed-inverse-limit-spectral-spaces}.
    Let $U \subset X$ be an open neighbourhood of $W$.
    Then $E_i = U_i \cap (X \setminus U)$ is a family of constructible
    subsets of the spectral space $Z = X \setminus U$
    with empty intersection. Using that the spectral topology on $Z$
    is quasi-compact (Lemma \ref{lemma-constructible-hausdorff-quasi-compact})
    we conclude from
    Lemma \ref{lemma-intersection-closed-in-quasi-compact}
    that $E_i = \emptyset$ for some $i$.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 0A31

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?