## Tag `0052`

Chapter 5: Topology > Section 5.9: Noetherian topological spaces

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

- Any subset of $X$ with the induced topology is Noetherian.
- The space $X$ has finitely many irreducible components.
- 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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