[MUSIC] We're going to discuss how to compute the canonical basis. But before doing so, let's introduce now the definition. We'll say that a set Q is preclosed, If the following holds. Whenever, We take a subset S of Q, a proper subset S of Q. And this S is itself preclosed. Then the closure of S must be in Q, so S double prime is a subset of Q. Well, this definition is as circular as the one of pseudo-closed sets, and again, it's not a problem. And actually, this definition looks very much like the one for pseudo-closed sets. The only difference is that we don't require S to be non-closed. So a closed set would also satisfy this definition. Because a closed set contains the closure of all its subsets, not only proper subsets, not only proper preclosed subsets. So it's easy to see that a set is preclosed, If, and only if, it's either closed or pseudo-closed. So this new definition is not really new, it's just a shortcut for saying closed or pseudo-closed. Note, also, that we can reformulate this definition. And say that a set Q is, Preclosed, If the following condition holds. Whenever we have set P, which is a proper subset of Q, and which is, at the same time, pseudo-closed, Then its closure must be contained in Q. So the difference is not that I use S here and P here. The difference is that here I'm talking about preclosed S, and here I'm talking only about psuedo-closed sets. But that's an obvious simplification, because every preclosed set is either pseudo-closed or closed. But if a set contains a closed set, then it must also contain its closure. Because its closure is just the set itself. All right, so what's interesting about these preclosed sets is that they form a closure system. Now, it's a different closure system. Not the one corresponding to the closure operator double prime, for which these sets are preclosed. So it's a closure system corresponding to a different closure operator, but it's still a closure system. Let's prove this. Well, what we have to prove is that whenever we have several preclosed sets, their intersection must also be preclosed. So the intersection, Of preclosed sets is preclosed. So a closure system is a system of sets that contains the universal set, and that it's closed on their intersection. Now we'll prove that preclosed sets are closed on their intersection. And of course, the universal set is preclosed. Because it is closed, no matter what our closure operator is. So let's prove this. Suppose we have several preclosed sets. Let's denote them Qi, and let Q be their intersection. So Q is the intersection of Qi's. Well, i ranges from one to something, doesn't matter what. Now suppose, That Q is different from all Qis. So Q, Is not equal to Qi, for all i. Well, this is not the case. If Q is equal to some Qi, then Q is of course preclosed, because all Qis are preclosed. That's what we assume, so every Qi is preclosed. Okay, now I assume that Q is different from all Qis, but we still want to show that it is preclosed. So let's look at the definition. For a Q to be preclosed, it must satisfy this definition. Now let's take an arbitrary S, an arbitrary preclosed S, Which is a subset of Q. A proper subset of Q. Well, S is a subset of Q, which is the intersection of Qis. So S is a subset of every Qi. Actually, proper subset of every Qi. Because every Qi is preclosed, and it contains S, it must contain its closure. So S double prime is also a subset of Qi for all i. But then, as double prime must be contained in the intersection of all Qis, And this intersection is Q. So we've shown that if we take a preclosed S, which is a proper subset of Q, then its closure must be a subset of Q as well. And this holds for every preclosed subset of Q, so Q must be preclosed itself. So if we take preclosed sets and compute their intersection, it's going to be preclosed. Therefore, the intersection of preclosed sets is always preclosed, and preclosed sets form a closure system. [MUSIC]