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 $X$.

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 $X$. $\square$

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