The dimension of real space is the number of coordinates needed to represent a point. A line has one dimension, a plane has two dimensions, the space around you has three, and so on. However, when the space is a reflection of certain algebraic structures, dimension takes on a whole new meaning. It is one less than the length of the longest chain of irreducible closed sets in the space. This seems confusing, but when you get to algebraic geometry, Rn will be given a new topology, such that the irreducible dimension of the space agrees with the local euclidean dimension. Trust me, it all works out.
A nonempty set S in a larger topology T is irreducible if every two nonempty open subsets of S intersect. It can't be "reduced". Of course S is given the subspace topology. Note that S is connected.
The space itself is irreducible if every two nonempty open sets intersect. If there is only one nonempty open set, i.e. the indiscrete topology, 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. Not so with other topologies.
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.
An irreducible set S might be open, or closed, or neither. Let C be the closure of an irreducible set S. Any open set that intersects C intersects S, and this intersection is open with respect to S. Two nonempty open sets in C produce nonempty open sets in S, which intersect. Therefore C is also irreducible. The closure of an irreducible set is irreducible.
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.
Imagine an ascending chain of irreducible sets, and let U be their union. If U is the only open set of U then we are done. Otherwise let O1 and O2 be open in U. Select points p1 and p2 in O1 and O2, and march up the chain until an irreducible set contains both p1 and p2. Within this set, O1 and O2 intersect in q, and the same holds in U. Therefore the union 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 closed set is called a component of the space. Since every point starts a chain, components cover the space.
Once again mathematicians are overloading their words. A component may be one of the spaces in a topological product, a connected subspace of a disconnected space, a path connected subspace of a disconnected space, or a maximal irreducible subspace. Sorry for the confusion.
If a space is hausdorff, and if a set contains two points, these points can be separated in disjoint open sets. Thus if a set is irreducible it must consist of a single point. These are the irreducible components of a hausdorff space. Thus the most interesting spaces, in this chapter, are not hausdorff.
If you've studied noetherian rings and modules, you're in good shape, except you have to stand on your head. Noetherian usually restricts the lengths of ascending chains, but in this context noetherian restricts the lengths of descending chains. A space or subspace is noetherian if every descending chain of closed sets is finite. You can't keep finding smaller and smaller closed sets.
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. This will become clear in the next chapter.
If the space has finitely many closed sets it is automatically noetherian. This is assured when the space itself is finite.
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. However, the converse is not true. In the plane (hausdorff), irreducible closed sets are points, and each chain has length 1, but ever shrinking disks about the origin produce an infinite descending chain of closed sets.
If T is noetherian, every subspace of T is compact. Let W be a subspace of T and consider an open cover for W. Take complements to find closed sets in W that have no common intersection. Let p1 be a point in the first closed set C1. Find C2 that does not contain p1, and let D2 = C1∩C2. If D2 is empty we have our finite subcover. Otherwise let p2 be a point in D2 and find C3 that misses p2. Let D3 = D2∩C3. Repeat this process, building a descending chain of closed sets D2 D3 D4 etc. This chain must terminate, giving a finite subcover, whence W is compact.
Assume T is noetherian, and suppose a closed set C is not the finite union of irreducible closed sets. Descend down the chain to find a minimal example of this. This set is not itself irreducible, so that it is the union of two properly smaller closed sets. Each is the finite union of irreducible sets by minimality, so combine these to represent the original set as a finite union.
Next, show the decomposition into irreducible components exists and is unique. Given a closed set C, write C as a finite union of irreducible closed sets. Work within the space C from here on out. Drive each of these irreducible sets up to a maximal irreducible 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. A1, irreducible, lies in the finite union of B1 through Bn, and as shown earlier, A1 lies in one of them, say B1. A1 is maximal, so B1 can't be larger. Thus B1 = A1. However, equal components have already been paired up. This is a contradiction, and that completes the proof.
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 that are contained in C.
If H is an irreducible set in C, its closure is a closed irreducible set, and this rises to a component of C. Every irreducible set in C belongs to at least one component of C.
The dimension of a space is the length of the longest chain of irreducible closed sets, minus one. The same definition applies to a subspace W of the larger space T. The chain of irreducible closed sets must lie in W, using the subspace topology.
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.
Let R be any region in a topological space T, and let W be a closed set in R, using the subspace topology. In other words, W is the intersection of R and a closed set in T. Since the closure of W, denoted W′, lies in R′, closure is a map from the closed sets in R to the closed sets in R′. If R is the open disk in the plane, and W is a slice of pie without the crust, then W′ is the complete slice of pie in the closed disk R′. Closure puts the boundary on the slice of pie. This map need not be onto. For instance, there may be closed sets in R′ that are outside of R. You can see this in euclidean space, where every point is closed, hence the boundary of an open set certainly contains closed sets. However, the map is injective, as shown below.
Intersect W′ with R and you get at least W back again. Suppose there is some x in W′ and in R that is not in W. Since W is closed, some open set contains x and misses W. This proves x cannot lie in W′ after all. Intersection with R maps W′ back to the W that created it. The closed sets in R embed, via closure, into the closed sets in R′.
Let W be irreducible and suppose W′ is the nontrivial union of closed sets U and V in R′. Intersect U and V with R and find two closed sets that combine to form W. One of these has to equal W. Let U and R intersect in W. Since U is a closed set it contains W′. We didn't need V after all. Therefore W′ is irreducible. Closure maps the irreducible closed sets of R into the irreducible closed sets of R′.
Since closure respects order, chains of irreducible closed sets in R map to chains of irreducible closed sets in R′. The dimension of R′ is at least the dimension of R. It could be greater, as shown in the following example.
Label the points of a space a through z. The sets a, ab, abc, abcd, abcde, … a through z are open, hence the trailing segments are closed. Each closed set is irreducible. Let the region R be the single point a, hence the closure of R is the whole space S. R has dimension 0, and S has dimension 25.
If finitely many closed sets cover R, their closures cover all of R′. If x in R′ is not covered, it lies in an open set, missing each of the closed sets of R. Intersect these open sets to find an open set that separates x from R. This contradicts x in the closure of R. Thus the closures provide a closed cover for R′.
Assume R′ is noetherian, hence every subspace, including R, is noetherian. Write R as a finite union of components and let G be one of these components. Now G′ is a closed irreducible set in R′. Push this up to a component H of R′. Suppose G belongs to two components H1 and H2. The other components of R lie, at least partly, outside of G. All these live in other components of R′, not H1 nor H2. The closures of the components of R cover R′, so in each case, we can choose any component we like in R′, containing each component of R, and the result covers R′. Since the finite component cover of R′ is unique, we can't really choose between H1 and H2. Therefore G belongs to one specific component H.
Remember that G′ is irreducible in R′, and so it belongs to just one component, which has to be H. Intersect H with R, which is the same as the intersection of G′ with R, which is G. Once again, intersection with R reverses the injective map from the components of R into the components of R′.
The components covering R′ are unique, so there can't be any more. The map is surjective, as well as injective. The components of R and the components of R′ correspond 1 for 1.
As it turns out, the component H that corresponds to G is precisely G′. Suppose H includes some extra stuff beyond G′. Intersect H with the union of the closures of the other components of R, omitting G, and call this set Z. Note that Z is closed in R′. The closures combine to cover all of R′, so Z contains everything in H-G′. If these closures include all of G, then intersect with R, and the union of the other components covers R, and we didn't need G after all. The closures do not cover all of G, nor all of G′. G′ and Z combine to form H, and H is reducible. This is a contradiction, hence the unique component associated with G is G′, and G′ intersects R to give G back again.
This is illustrated by our earlier alphabet example. The component of R is a, and the component of R′ is the entire alphabet. These correspond through closure and intersection.
Returning to real space, let R be the open disk in the plane. Points are irreducible components, and they remain so in R′. However, the closed disk brings in more points on the boundary, more components that don't correspond to any components of R. There is no contradiction here, since R′ is not noetherian.
In any space, the closure of a point x is an irreducible closed set. The closure of x is certainly closed, and any open set that intersects the closure contains x, whence any two open sets intersect in x, and the closure becomes irreducible.
Given a closed irreducible set C, x is a generic point for C if the closure of x yields C.
If every closed irreducible set has a generic point, the space is generic.
Any space S can be transformed into a generic space by building a space T, where the points of T are the irreducible closed sets of S. If U is closed in S, let the collection of closed irreducible sets in U be closed in T. Set U = S and T remains closed, as it should. The empty set is also closed. Verify that the arbitrary intersection and finite union of closed sets in S implies the same for T. Intersection is easy; union takes a bit more thought. Suppose U and V are closed in S, and their union brings in an irreducible closed set that is not in U or V alone. This builds a nontrivial cover for the irreducible set, which is a contradiction. Therefore the topology is valid.
Let a function f map S into T by taking any x to its closure, which is an irreducible closed set. Thus points of S map to points of T.
The preimage of a closed set in T is the collection of points whose closures all live in a closed set C in S. x satisfies this criterion iff x lives in C, hence the preimage is closed, and f is continuous.
Conversely, start with a closed set C in S. The closure of every point in C remains in C, hence the image in T is closed, and f is bicontinuous. To be technically correct, the function f, from S onto its image in T, using the subspace topology inherited from T, is bicontinuous. After all, there may be points of T that don't pull back to anything in S.
Assume x and y have the same closure, and map to the same point in T. Every open set containing x contains y, and every open set containing y contains x. Turn this around and express it this way. If S is half hausdorff, or T0, then f is an embedding. In fact f is a bicontinuous embedding, which means S is a subspace of T. Putting S into its generic space sort of fills in the gaps. The topology of S, within T, has not changed.
If S is hausdorff, each point is already closed and irreducible, and these are the only irreducible sets, hence S and T are homeomorphic.
We still need to prove T is generic. Let U be closed in S, with some point x that is not in a closed set V. The closure of x is a point in T that is in the closed set defined by U, but not the closed set defined by V. If a closed set in S is reducible, then the corresponding closed set in T is reducible. Turn this around and an irreducible closed set in T comes from an irreducible closed set in S.
Let W be an irreducible closed set in T and let U be the corresponding set in S. Since U is irreducible, the set in T that is defined by U includes a point q that corresponds to U. Any closed set in T containing q comes from a closed set in S containing U. This brings in all of W, hence W is the closure of q. Every closed irreducible set in T is generic, and T is a generic space.
Let q and r be two points in T, thus two irreducible closed sets in S. They can't contain each other in S, so one defines a closed set in T that does not include the other. Thus T is half hausdorff.
What if we try to generize T, or any other space that is already generic and half hausdorff? Let the points of Y be the irreducible closed sets of T. Each point in Y corresponds to an irreducible set in T, which (being a generic space) is the closure of some point x in T. Thus T maps onto Y. With T half hausdorff, the map is injective. If a space T is half hausdorff and generic, the transformation of T is a homeomorphism, and doesn't change T at all. There is no need to generize a space twice.
The generizing function f is bicontinuous, which means closed sets correspond. Furthermore, the preimage of a reducible set is reducible, hence the image of an irreducible set is irreducible. Run this in both directions and irreducible closed sets in S and T correspond. This makes sense, since S is usually a subspace of T. f preserves the order of containment, hence S is noetherian iff T is noetherian, and the dimension of S equals the dimension of T.
A zariski space T is generic, half hausdorff, and noetherian. There is no need to generize T, as described in the previous section, since the result would be homeomorphic to T.
If T is the generic space produced from any other noetherian space S, then T is generic, half hausdorff, and noetherian, hence zariski.
These spaces are named after Oscar Zariski, a Russian mathematician who placed a noetherian topology on top of various rings and vector spaces, in order to answer certain algebraic questions. The real power is noetherian, since such a space can always be generized later, mostly for convenience. This will become clearer in the next section. It's a wonderful blending of algebra and topology, and it won't be the last.
By dcc, every closed set in T has a minimal closed subset, which must be irreducible. Call this minimal set C, and let x be its generic point. Since T is half hausdorff, there can be no other points in C. Thus the minimal closed sets are points. These are the closed points of T.
If y ≠ x, and y is in the closure of x, we say x specializes to y, and y generizes to x. This is not a function, rather, it induces a partial ordering on the points of T. Let x > y indicate x specializes to y, or y is in the closure of x. If z is in the closure of y, then the closure of x brings in y, brings in z, so the relation is transitive. If x is in the closure of y and y is in the closure of x, we are violating half hausdorff, hence the relation is antisymmetric. Thus it establishes a partial ordering.
If x is minimal then its closure brings in nothing else. It is closed and irreducible - one of the closed points. Conversely let x be a closed point in T. It's closure brings in only x, and x is minimal.
Assume x is maximal, and let C be its closure, an irreducible set. If C is contained in a larger irreducible set, with generic point y, then the closure of y includes x, which is a contradiction. The maximal points correspond to the irreducible components of T.
If x is in a closed set V, the closure of x lies in V. This holds for all x in V. V is stable under specialization.
If U is open and U misses x, U misses a closed set containing x, and U misses the closure of x. Equivalently, if U contains the closure of x it contains x. This means U is stable under generization.