1. Noetherian Space
Let X be a topological space. We say that X is irreducible if X = X1∪ X2, where X1 and X2 are closed, then either X = X1 or X = X2. The empty set is not considered to be irreducible. A subset Y of a topological space X is irreducible if Y is irreducible when we equip Y with the subspace topology.
Definition 1.1. A topological space is Noetherian if it satisfies the descending chain con- ditions on the closed sets: for any sequence of closed sets Y1 ⊃ Y2 ⊃ · · · there exists r ≥ 1 so that Yi = Yr for all i ≥ r.
Theorem 1.1. In a noetherian space X, every nonempty closed subset Y can be expressed as a finite union Y = Y1∪ · · · Yr of irreducible closed subsets Yi. If we require that Yi 6⊇ Yj for i 6= j, then Yi are uniquely determined. They are called the irreducible components of Y.
Proof. Claim: every closed subset Y of X has such a decomposition.
Let S be a set of all closed subsets of X which can not be decomposed into a finite union of irreducible closed subsets. If we can show that S is an empty set, then we prove our claim.
Assume that S is nonempty. Let us define a relation ≤ on S by A ≤ B if A ⊃ B. Suppose {Cn} is a chain in S: C1 ≤ C2 ≤ C3 ≤ · · · . Then C1 ⊃ C2 ⊃ · · · . Since X is a noetherian space, there is r > 0 so that Ci = Crfor all i ≥ r. Take C = Cr. Then we know that C ∈ S.
Since Ci ≤ C = Cr for all i, C is an upper bound of the chain (Cn). Since every chain in (S, ≤) has an upper bound, by the Zorn’s Lemma, we can find Y so that Y is maximal in (S, ≤). Since Y ∈ S, Y is not irreducible. Assume that Y = Y1∪ Y2, where Y1 and Y2
are proper closed subsets of Y. Then Y ≤ Y1 and Y ≤ Y2. Since Y is in S, Y is not a finite union of irreducible closed subsets. Hence either Y1 or Y2 are not irreducible. If Y1
is not irreducible, Y1 ∈ S. Since Y is maximal in S and Y ≤ Y1, Y = Y1 which leads to a contradiction that Y1 is a proper subset of Y. Therefore S must be empty.
Let us assume that Y = Y1∪ · · · ∪ Yr and Y = Y10∪ · · · ∪ Ys0 are two representations of Y satisfying the required properties. Then Y10 ⊂ Y implies that Y10 =Sr
i=1(Y10∩ Yi). Since Y10 is irreducible, Y10∩ Yi = Y10 for some i. Say i = 1. Then Y10 ⊂ Y1. Similarly, let us consider Y1 ⊂ Y10∪ · · · ∪ Yr0, and then obtain that Y1 ⊂ Yj0 for some j. Since Y10 ⊂ Y1, we find Y10 ⊂ Yj0. By the assumption (we assume that Yi0 6⊇ Yj0 for i 6= j,), we find j = 1, i.e. Y1 ⊂ Y10. Hence Y1 = Y10. Let Z be the closure of Y − Y1. Then Z = Y2∪ · · · ∪ Ys= Y20∪ · · · ∪ Yr0. Inductively, we can show that (after rearranging the indices of Yi) Yi = Yi0 and r = s. A topological space X is called quasi compact if every open cover of X has a finite sub cover. (Usually, such a space is called compact. In algebraic geometry, we call such a space quasi compact.)
Definition 1.2. Let X be a topological space. We define the dimension of X, denoted by dim X, to be the supremum of all integers n such that there exists a chain Z0 ⊂ · · · ⊂ Zn of distinct irreducible closed subsets of X.
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