The Stacks project

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 (Lemma 5.8.4), 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 $X$. $\square$

Proof. In case (1), suppose that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots $ is a descending chain of closed subsets of $f(X)$ (as usual with the induced topology as a subset of $Y$). Then $f^{-1}(Z_1) \supset f^{-1}(Z_2) \supset f^{-1}(Z_3) \supset \ldots $ is a descending chain of closed subsets of $X$. Hence this chain stabilizes. Since $f(f^{-1}(Z_ i)) = Z_ i$ we conclude that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots $ stabilizes also. In case (2), let $y \in f(X)$. Choose $x \in X$ with $f(x) = y$. By assumption there exists a neighbourhood $E \subset X$ of $x$ which is Noetherian. Then $f(E) \subset f(X)$ is a neighbourhood which is Noetherian by part (1). $\square$

Proof. Let $\{ F_ m\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X' = \bigcup _{i = 1, \ldots , n} X_ i$ with the induced topology. Then we can find a decreasing sequence $\{ G_ m\} _{m \in \mathbf{N}}$ of closed subsets of $X$ verifying $F_ m = G_ m \cap X'$ for all $m$ (small detail omitted). As $X_ i$ is noetherian and $\{ G_ m \cap X_ i\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X_ i$, there exists $m_ i \in \mathbf{N}$ such that for all $m \geq m_ i$ we have $G_ m \cap X_ i = G_{m_ i} \cap X_ i$. Let $m_0 = \max _{i = 1, \ldots , n} m_ i$. Then clearly

stabilizes for $m \geq m_0$ and the proof is complete. $\square$

Proof. Let $x \in X$. Let $E$ be a neighbourhood of $x$. We have to find a connected neighbourhood of $x$ contained in $E$. By assumption there exists a neighbourhood $E'$ of $x$ which is Noetherian. Then $E \cap E'$ is Noetherian, see Lemma 5.9.2. Let $E \cap E' = Y_1 \cup \ldots \cup Y_ n$ be the decomposition into irreducible components, see Lemma 5.9.2. Let $E'' = \bigcup _{x \in Y_ i} Y_ i$. This is a connected subset of $E \cap E'$ containing $x$. It contains the open $E \cap E' \setminus (\bigcup _{x \not\in Y_ i} Y_ i)$ of $E \cap E'$ and hence it is a neighbourhood of $x$ in $X$. This proves the lemma. $\square$