Noetherian topological space

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A topological space $X$ is called {\PMlinkescapetext {\em Noetherian}} if it satisfies the descending chain condition for closed subsets: for any sequence
$Y_1 \supseteq Y_2 \supseteq \cdots$
of closed subsets $Y_i$ of $X$, there is an integer $m$ such that
$Y_m=Y_{m+1}=\cdots$.

As a first example, note that all finite topological spaces are Noetherian.

There is a lot of interplay between the Noetherian condition and compactness:
\begin{itemize}
\item Every Noetherian topological space is quasi-compact.
\item A Hausdorff topological space $X$ is Noetherian if and only if every subspace of $X$ is compact. (i.e. $X$ is hereditarily compact)
\end{itemize}

Note that if $R$ is a Noetherian ring, then $\text{Spec}(R)$, the prime spectrum of $R$, is a Noetherian topological space.

{\bf Example of a Noetherian topological space:}\\
The space $\mathbb{A}^n_k$ (affine $n$-space over a field $k$) under the Zariski topology is an example of a Noetherian topological space.  By properties of the ideal of a subset of $\mathbb{A}^n_k$, we know that if
$Y_1 \supseteq Y_2 \supseteq \cdots$ is a descending chain of Zariski-closed subsets, then $I(Y_1) \subseteq I(Y_2) \subseteq \cdots$ is an ascending chain of ideals of $k[x_1,\ldots,x_n]$.

Since $k[x_1,\ldots,x_n]$ is a Noetherian ring, there exists an integer $m$ such that $I(Y_m)=I(Y_{m+1})=\cdots$.  But because we have a one-to-one correspondence between radical ideals of $k[x_1,\ldots,x_n]$ and Zariski-closed sets in $\mathbb{A}^n_k$, we have $V(I(Y_i))=Y_i$ for all $i$.  Hence
$Y_m=Y_{m+1}=\cdots$ as required.
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