Irreducible and Noetherian topological spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 2 June 2011

Noetherian spaces

A non-empty topological space $X$ is irreducible if every pair of non-empty open sets in $X$ intersect (thus $X$ is as far as possible from being Hausdorff). Equivalent conditions:

(a)   $X$ is not the union of two proper closed subsets.

(b)   If $F_i$ ($1\le i\le n$) are closed subsets which cover $X$, then $X=F_i$ for some $i$.

(c)   Every non-empty open set is dense in $X$.

(d)   Every open set in $X$ is connected.

Examples.

(1)   Let $X$ be an infinite set, and topologize $X$ by taking the closed subsets to be $X$ itself and all finite subsets of $X$. Then $X$ is irreducible.

(2)   Any irreducible algebraic variety with the Zariski topology.

A subset $Y$ of a space $X$ is irreducible if $Y$ is irreducible in the induced topology. The following facts are not hard to prove:

(i)   If ${\left(F_i\right)}_{1\le i\le n}$ is a finite closed covering of a space $X$, and if $Y$ is an irreducible subset of $X$, then $Y\subseteq F_i$ for some $i$.

(ii)   If $X$ is irreducible, every non-empty open subset of $X$ is irreducible.

(iii)   Let ${\left(U_i\right)}_{1\le i\le n}$ be a finite open covering of a space $X$, the $U_i$ being non-empty. Then $X$ is irreducible if and only if each $Ui$ is irrducible and meets each $U_j$.

(iv)   If $Y$ is a subset of $X$, then $Y$ is irreducible if and only if $\bar{Y}$ is irreducible.

(v)   The image of an irreducible set under a continuous map is irreducible.

(vi)   $X$ has maximal irreducible subsets; they are all closed and they cover $X$. (Use Zorn's lemma for (vi).)

The maximal irreducible substes of $X$ are called the irreducible components of $X$. Irreducibility is in some ways analogous to, but stronger than, connectedness.

If $x\in X$, then $\left\{x\right\}$ is irreducible and therefore (by (iv) above) so is $\overline{\left\{x\right\}}$. If $V$ is an irreducible subset of $X$ and $V=\overline{\left\{x\right\}}$ for some $x\in X$, then $x$ is a generic point of $V$. If $y\in \overline{\left\{x\right\}}$, $y$ is a specialization of $x$. The closed set $\overline{\left\{x\right\}}$ is the locus of $x$.

A subset $Y$ of a space $X$ is locally closed if $Y$ is the intersection of an open set and a closed set in $X$, or equivalently if $Y$ is open in its closure $\bar{Y}$, or equivalently again if every $y\in Y$ has an open neighborhood $U_y$ in $X$ such that $Y\cap U_y$ is closed in $U_y$.

A topological space is Noetherian if the closed subsets of $X$ satisfy the descending chain condition. Equivalent conditions:

• The open sets in $X$ satisfy the ascending chain condition;
• Every open subset of $X$ is quasi-compact (i.e. compact but not necessarily Hausdorff);
• Every subset of $X$ is quasi-compact.

(i)   A Noetherian space is quasi-compact.

(ii)   Every subset of a Noetherian space (with the induced topology) is Noetherian.

(iii)   Let $X$ be a topological space and let ${\left(X_i\right)}_{1\le i\le n}$ be a finite covering of $X$. If the $X_i$ are Noetherian, then so is $X$.

(iv)   If $X$ is Noetherian, the number of irreducible components of $X$ is finite.

The proofs are straightforward.

Notes and References

These notes are taken from [Mac].

References

[Mac] I.G. Macdonald, Algebraic Geometry: Introduction to Schemes, W.A. Benjamin, New York, 1968.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

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