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Tag 0052

Chapter 5: Topology > Section 5.9: Noetherian topological spaces

Lemma 5.9.2. Let $X$ be a Noetherian topological space.

  1. Any subset of $X$ with the induced topology is Noetherian.
  2. The space $X$ has finitely many irreducible components.
  3. Each irreducible component of $X$ contains a nonempty open of $X$.

Proof. Let $T \subset X$ be a subset of $X$. Let $T_1 \supset T_2 \supset \ldots$ be a descending chain of closed subsets of $T$. Write $T_i = T \cap Z_i$ with $Z_i \subset X$ closed. Consider the descending chain of closed subsets $Z_1 \supset Z_1\cap Z_2 \supset Z_1 \cap Z_2 \cap Z_3 \ldots$ This stabilizes by assumption and hence the original sequence of $T_i$ stabilizes. Thus $T$ is Noetherian.

Let $A$ be the set of closed subsets of $X$ which do not have finitely many irreducible components. Assume that $A$ is not empty to arrive at a contradiction. The set $A$ is partially ordered by inclusion: $\alpha \leq \alpha' \Leftrightarrow Z_{\alpha} \subset Z_{\alpha'}$. By the descending chain condition we may find a smallest element of $A$, say $Z$. As $Z$ is not a finite union of irreducible components, it is not irreducible. Hence we can write $Z = Z' \cup Z''$ and both are strictly smaller closed subsets. By construction $Z' = \bigcup Z'_i$ and $Z'' = \bigcup Z''_j$ are finite unions of their irreducible components. Hence $Z = \bigcup Z'_i \cup \bigcup Z''_j$ is a finite union of irreducible closed subsets. After removing redundant members of this expression, this will be the decomposition of $Z$ into its irreducible components, a contradiction.

Let $Z \subset X$ be an irreducible component of $X$. Let $Z_1, \ldots, Z_n$ be the other irreducible components of $X$. Consider $U = Z \setminus (Z_1\cup\ldots\cup Z_n)$. This is not empty since otherwise the irreducible space $Z$ would be contained in one of the other $Z_i$. Because $X = Z \cup Z_1 \cup \ldots Z_n$ (see Lemma 5.8.3), also $U = X \setminus (Z_1\cup\ldots\cup Z_n)$ and hence open in $X$. Thus $Z$ contains a nonempty open of $X$. $\square$

    The code snippet corresponding to this tag is a part of the file topology.tex and is located in lines 1014–1022 (see updates for more information).

    \begin{lemma}
    \label{lemma-Noetherian}
    Let $X$ be a Noetherian topological space.
    \begin{enumerate}
    \item Any subset of $X$ with the induced topology is Noetherian.
    \item The space $X$ has finitely many irreducible components.
    \item Each irreducible component of $X$ contains a nonempty open of $X$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Let $T \subset X$ be a subset of $X$.
    Let $T_1 \supset T_2 \supset \ldots$
    be a descending chain of closed subsets of $T$.
    Write $T_i =  T \cap Z_i$ with $Z_i \subset X$ closed.
    Consider the descending chain of closed subsets
    $Z_1 \supset Z_1\cap Z_2 \supset Z_1 \cap Z_2 \cap Z_3 \ldots$
    This stabilizes by assumption and hence the original sequence
    of $T_i$ stabilizes. Thus $T$ is Noetherian.
    
    \medskip\noindent
    Let $A$ be the set of closed subsets of $X$ which do not
    have finitely many irreducible components. Assume that
    $A$ is not empty to arrive at a contradiction.
    The set $A$ is partially ordered by inclusion: $\alpha \leq \alpha'
    \Leftrightarrow Z_{\alpha} \subset Z_{\alpha'}$.
    By the descending chain condition we may find a
    smallest element of $A$, say $Z$. As $Z$ is not a finite
    union of irreducible components, it is not irreducible.
    Hence we can write $Z = Z' \cup Z''$ and both are strictly smaller
    closed subsets. By construction $Z' = \bigcup Z'_i$ and
    $Z'' = \bigcup Z''_j$ are finite unions of their irreducible
    components. Hence $Z = \bigcup Z'_i \cup \bigcup Z''_j$ is
    a finite union of irreducible closed subsets.
    After removing redundant members of this expression,
    this will be the decomposition of $Z$ into its irreducible
    components, a contradiction.
    
    \medskip\noindent
    Let $Z \subset X$ be an irreducible component of $X$.
    Let $Z_1, \ldots, Z_n$ be the other irreducible components
    of $X$. Consider $U = Z \setminus (Z_1\cup\ldots\cup Z_n)$.
    This is not empty since otherwise the irreducible space
    $Z$ would be contained in one of the other $Z_i$.
    Because $X = Z \cup Z_1 \cup \ldots Z_n$ (see Lemma \ref{lemma-irreducible}),
    also $U = X \setminus (Z_1\cup\ldots\cup Z_n)$
    and hence open in $X$. Thus $Z$ contains a nonempty
    open of $X$.
    \end{proof}

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