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Tag 01YY

Chapter 31: Limits of Schemes > Section 31.4: Descending properties

Lemma 31.4.6. In Situation 31.4.5.

  1. We have $S_{set} = \mathop{\rm lim}\nolimits_i S_{i, set}$ where $S_{set}$ indicates the underlying set of the scheme $S$.
  2. We have $S_{top} = \mathop{\rm lim}\nolimits_i S_{i, top}$ where $S_{top}$ indicates the underlying topological space of the scheme $S$.
  3. If $s, s' \in S$ and $s'$ is not a specialization of $s$ then for some $i \in I$ the image $s'_i \in S_i$ of $s'$ is not a specialization of the image $s_i \in S_i$ of $s$.
  4. Add more easy facts on topology of $S$ here. (Requirement: whatever is added should be easy in the affine case.)

Proof. Part (1) is a special case of Lemma 31.4.1.

Part (2) is a special case of Lemma 31.4.2.

Part (3) is a special case of Lemma 31.4.4. $\square$

    The code snippet corresponding to this tag is a part of the file limits.tex and is located in lines 391–405 (see updates for more information).

    \begin{lemma}
    \label{lemma-topology-limit}
    In Situation \ref{situation-descent}.
    \begin{enumerate}
    \item We have $S_{set} = \lim_i S_{i, set}$ where $S_{set}$
    indicates the underlying set of the scheme $S$.
    \item We have $S_{top} = \lim_i S_{i, top}$ where $S_{top}$
    indicates the underlying topological space of the scheme $S$.
    \item If $s, s' \in S$ and $s'$ is not a specialization of $s$
    then for some $i \in I$ the image $s'_i \in S_i$ of $s'$ is not
    a specialization of the image $s_i \in S_i$ of $s$.
    \item Add more easy facts on topology of $S$ here.
    (Requirement: whatever is added should be easy in the affine case.)
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Part (1) is a special case of Lemma \ref{lemma-inverse-limit-sets}.
    
    \medskip\noindent
    Part (2) is a special case of Lemma \ref{lemma-inverse-limit-top}.
    
    \medskip\noindent
    Part (3) is a special case of Lemma \ref{lemma-inverse-limit-irreducibles}.
    \end{proof}

    Comments (2)

    Comment #1254 by Michael on January 9, 2015 a 11:35 am UTC

    there is a typo in the proof of (2): deisred should be desired

    Comment #1265 by Johan (site) on January 22, 2015 a 7:41 pm UTC

    OK, thanks! Fixed here.

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