irreducible closed subspace

Given a topological space $X$, a closed subspace $F$ of $X$ is **irreducible** if there are exactly two ways to express $F$ as a union of two closed subspaces: $F = \empty \cup F$ and $F = F \cup \empty$. In other words, $F$ must be inhabited (so that these two ways are distinct) but it must be impossible to express $F$ as a union of two inhabited closed subspaces.

Note that the closure of any point of $X$ is an irreducible closed subspace. $X$ is sober if and only if every irreducible closed subspace is the closure of a unique point of $X$. In general, the irreducible closed subspaces of $X$ correspond to the points of the topological locale $\Omega(X)$.

The theory of irreducible closed subspaces is not useful in constructive mathematics; instead, one should use the completely prime filters on the frame of open subspaces of $X$ (which are the points of $\Omega(X)$ *by definition*).

Created on July 1, 2010 17:25:43
by Toby Bartels
(98.16.139.29)