Last updates: 2 June 2011

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.
- (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.
- (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 $\stackrel{\u203e}{Y}$ 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 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*.

**Examples.**

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:

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
$\stackrel{\u203e}{\left\{x\right\}}$.
If $V$ is an irreducible subset of $X$ and
$V=\stackrel{\u203e}{\left\{x\right\}}$ for some $x\in X$, then $x$
is a **generic point** of $V$. If
$y\in \stackrel{\u203e}{\left\{x\right\}}$, $y$ is a **specialization** of $x$.
The closed set
$\stackrel{\u203e}{\left\{x\right\}}$
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 $\stackrel{\u203e}{Y}$, 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 proofs are straightforward.

These notes are taken from [Mac].

[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.