applet-magic.com
Thayer Watkins
Silicon Valley
& TornadoAlley
USA

Kuratowski's Closure Operation in Topology

A topology for a set S is a collection of subsets of S such that

The elements of the collection are called the open sets of the topology. It is important to recognize that the openness of a set is not a property of the set itself; openness refers only to the membership of the set in the collection of subsets which is called the topology.

A set is defined as being closed with respect to a topology if its complement is open with respect to the topology; i.e., if its complement belongs to the topology. At least two sets, the null set and the whole set S, are both open and closed in any topology of S.

If the closed sets of a topology are given the open sets can easily be constructed since they are simply the complements of the closed sets.

The Polish mathematician Kazimierz (Casimir) Kuratowski (1896-1980) developed a radically different approach to specifying a topology for a set. Kuratowski considered particular functions from the set of subsets of S to the set of subsets of S; i.e.,


K:P(S)→P(S)
where P(S) is the set of all subsets of S,
the so-called power set of S
 

which satisfy the conditions

These conditions imply a number of other properties of a closure function.


Lemma K0: For any nonempty subset X of S, X ⊆ K(X).
 


Proof: Consider any element x of X. Since X=(X-{x})∪{x} it follows that x is a member of K(X) because K(X)=K((X-{x})∪K({x}) = K((X-{x}))∪{x}. Thus X ⊆ K(X).

The null set is a special case. By assumption ∅=K(∅) so ∅ ⊆ K(∅).



Lemma k1: For any two subsets X and Y such that X ⊆ Y then K(X) ⊆ K(Y).
 


Proof: If X ⊆ Y then X∪Y = Y and thus K(X∪Y) = K(X)∪K(Y) = K(Y) which implies that K(X) ⊆ K(Y).



Lemma k2: For any subsets X and Y, K(X∩Y) ⊆ K(X)∩K(Y).
 


The intersection X∪Y ⊆ is a subset of X and a subset of Y. By Lemma 1 this means K(X∪Y) is a subset of K(X) and a subset of K(Y). Thus K(X∪Y) is a subset of K(X)⊆K(Y).



Lemma k3: For any two subsets X and Y, (K(X)-K(Y)) ⊆ K(X-Y).
 


Proof: Since X=(X-Y)∪(X∩Y) and hence K(X)=K(X-Y)∪K(X∩Y). Let x be any element of K(X) that does not belong to K(Y). The element x cannot belong to X∩Y because then by Lemma K1 it would belong to K(Y). It does not belong to K(Y)∩K(X). Since by Lemma k1 K(X∩Y) is a subset of K(X)∩K(Y) any element of K(X) that does not belong to k(Y) does not belong to K(X∩Y) and hence necessarily belongs to K(X-Y). Thus (K(X)-K(Y)) ⊆ K(X-Y).


Consider the collection L = {X: X⊂S and K(X)=X}. It is to be shown that this collection constitutes a collection of closed sets in which arbitrary intersection, finite unions, the null set ∅, and the whole set S belong to the collection.


Lemma K4: If K(Xα)=Xα for α∈I then K(∩Xα) = ∩K(Xα)
 


Proof: For each β∈I, ∩Xα ⊆ Xβ. Therefore ∩Xα ⊆ K(Xβ) for each β and hence ∩Xα ⊆ ∩K(Xβ)

Since K(Xβ)=Xβ this means that ∩Xα ⊆ ∩K(Xβ). But by Lemma K0, ∩Xβ ⊆ K(∩Xβ). These two relationships imply that


K(∩Xβ) = ∩Xβ
 

and thus ∩Xβ belongs to L.



Lemma K5: If K(X)=X and K(Y)=Y then K(X∪Y)=X∪Y.
 


Proof: K(X∪Y) = K(X)∪K(Y), but K(X) is the same as X and K(Y) is the same as Y so K(X∪Y)=X∪Y.

This means that if X and Y belong to L then X∪Y also belongs to L. This can be extended to any finite union.


Clearly ∅=K(∅) and S=K(S) belong to the collection L.

Therefore


Theorem K1: A Kuratowski closure operation on the power set of a set S generates a collection of subsets of S which satisfy the closed set conditions and thus establish a topology for the set S.
 


(To be continued.)

HOME PAGE OF Thayer Watkins
HOME PAGE OF applet-magic.com