The space itself is irreducible if every two nonempty open sets intersect. If there is only one nonempty open set, the space is irreducible by default.
Irreducible spaces are strange indeed. If you think about the plane, there are lots of open sets that don't intersect. Just take an open disk here and an open disk there. Even the tiniest region in the plane is not irreducible, because we can always find two distinct open disks inside this region.
It is often convenient to work with closed sets. A space is reducible iff two nonempty open sets do not intersect, iff two proper closed sets cover the space. If finitely many proper closed sets cover the space, bring them in one at a time until the space is covered. If the nth set does the trick, then the union of the first n-1 sets, and the nth set, prove the space is reducible.
If an irreducible set S lies in the finite union of closed sets, then it lies entirely in one of them. Suppose not, and intersect S with each of the closed sets in turn. The result is a collection of proper closed subsets of S that cover S. This contradicts the previous paragraph, hence S belongs to one of them.
Going back down, a nonempty open subset of an irreducible set is irreducible. Let C be an irreducible set, and let O be an open subset of C. Suppose A and B are nonempty disjoint open subsets of O. These come from open subsets of C that must intersect; but they must intersect outside of O. Intersect A with O, and B with O, and find open subsets of C that do not intersect. This is a contradiction, hence O is irreducible.
a point is an irreducible set, which starts the chain. Use zorn's lemma to assert the existence of a maximal irreducible set. Such a set is closed, else we could take its closure. Each maximal irreducible set is called an irreducible component of the space. Since every point starts a chain, the components cover the space.
If a space is hausdorff, and if a set contains two points, these points can be separated in open sets. Thus if a set is irreducible it must consist of a single point. These are the irreducible components of a hausdorff space.
Once again mathematicians are overloading their words. A component may be the projection of a product space, an open and closed (disconnected) subspace, or a maximal irreducible subspace. Try and guess from the context, and good luck!
Why the switch? Well - these spaces usually correspond to algebraic structures such as rings and modules, and the correspondence tends to be upside down. Thus smaller closed sets correspond to larger modules or ideals, and noetherian spaces come from noetherian rings. That's the rationale. If you want to check it out, have a look at zariski structures, but not until you've finished this introductory page. Now - back to noetherian spaces.
If the space has finitely many closed sets it is automatically noetherian. This is assured when the space itself is finite, i.e. consists of finitely many points.
Let W be a closed subspace of a noetherian space T. If W has an infinite descending chain of closed sets it defines an infinite chain of closed sets in T. This is a contradiction, hence W is noetherian. A closed subspace of a noetherian space is noetherian.
What if W is not a closed set? Suppose W has an infinite descending chain of closed sets within the topology of W. Thus each set in the chain comes from a closed set in T intersected with W. Let U and V be the first two sets in T that begin this descending chain. Replace V with V intersect U. We know that V was properly contained in U when restricted to W, now V is inside U relative to T. Do this all the way down the line and build an infinite descending chain in T, which is a contradiction. Every subspace of T is noetherian.
If descending chains of closed sets are finite then the same is true for irreducible closed sets. I don't know about the converse; perhaps somebody could help me out here.
Next, show the decomposition into irreducible components exists and is unique. Given a closed set C, write C as the finite union of irreducible closed sets. Drive each of these up to a component, and C becomes a finite union of components.
Now suppose C can be written as the finite union of components in two different ways. Some of the components may be the same; set these to the side. suppose A1, a component in the first representation, is not equal to any of the components in the second representation. Intersect A1 with B1 B2 B3 … Bn, giving n different intersections. The component A1 has been cut into pieces. Remember that A1 is irreducible, so at most one of these intersections produces a proper subset of A1. All the others yield nothing, or A1 in its entirety. Now A1 is maximal, so if B1∩A1 = A1, then B1 = A1. We've ruled this out, hence there is only one component in the second representation, say B1, that intersects A1, and that yields a proper subset, so some of A1 is not covered, hence some of C is not covered. This is a contradiction, hence the irreducible components are the same in both representations.
If G is a component of C that is not part of the decomposition, toss it in to create a larger decomposition. This is a contradiction, hence the decomposition of C is precisely the components of C.
If H is an irreducible set, its closure is a closed irreducible set, and this rises to a component of C. Thus every irreducible set in C belongs to at least one component of C.
The dimension of a hausdorff space is 0.
Let W be a closed set in T. Cover W with irreducible components, and let m be the maximum dimension of these components. The chain of length m+1 proves W has dimension at least m. Suppose W has dimension greater than m. A chain proves the dimension, and this chain has a maximal set R at the top, which is irreducible and closed. If R is wholly contained in one of our components then the dimension of R cannot exceed m. So R is split among several components. This contradicts the irreducibility of R. Therefore the dimension of W is the largest dimension of any of its components.
Take the union of V and W (closed) and cover it with the components of V and the components of W. This shows the dimension of V∪W is the larger of the dimensions of V and W.
If V is a subset of W, so that V∪W = W, the dimension of V is bounded by the dimension of W.
It is easy to build a space with an infinite dimension. Start with the unit interval and let the subintervals from 0 to 1/n be closed. (Zero is closed by intersection.) Verify that these are all irreducible. This has an infinite dimension. Include only some of the closed intervals for a finite dimension.
Let T be the disjoint union of spaces with dimensions 1, 2, 3, 4, etc. Let a set be closed in T if it is equal to T, or the finite union of closed sets in the component spaces. Verify that any closed set in T, other than T, has finitely many closed sets below it. A descending chain starts at T, moves to one of these closed sets, and then descends through finitely many closed sets. Therefore T is noetherian, yet its dimension is infinite.