03-top-constr

03-top-constr - Lecture 3 3. Constructing topologies In...

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Unformatted text preview: Lecture 3 3. Constructing topologies In this section we discuss several methods for constructing topologies on a given set. Definition. If T and T are two topologies on the same space X , such that T T (as sets), then T is said to be stronger than T . Equivalently, we will say that T is weaker than T . Remark that this condition is equivalent to the continuity of the map Id : ( X, T ) ( X, T ) . Comment. Given a (non-empty) set X , and a collection S of subsets of X , one can ask the following: Question 1 : Is there a topology on X with respect to which all the sets in S are open? Of course, this question has an affirmative answer, since we can take as the topology the collection of all subsets of X . Therefore the above question is more meaningful if stated as: Question 2 : Is there the weakest topology on X with respect to which all the sets in S are open? The answer to this question is again affirmative, and it is based on the following: Remark 3.1. If X is a non-empty set, and ( T i ) i I is a family of topologies on X , then the intersection i I T i is again a topology on X . In particular, if one starts with an arbitrary family S of subsets of X , and if we take ( S ) = T : T topology on X with T S , then the intersection top ( S ) = T ( S ) T is the weakest (i.e. smallest) among all topologies with respect to which all sets in S are open. The topology top ( S ) defined above cane also be described constructively as follows. Proposition 3.1. Let S be a collection of subsets of X . Then the sets in top ( S ) , which are a proper subsets of X , are those which can be written a (arbitrary) unions of finite intersections of sets in S . 13 14 LECTURE 3 Proof. It is useful to introduce the following notations. First we define V ( S ) to be the collection of all sets which are finite intersections of sets in S . In other words, B V ( S ) D 1 ,...,D n S such that D 1 D n = B. With the above notation, what we need to prove is that for a set A X , we have A top ( S ) V A V ( S ) such that A = [ B V A B. The implication is pretty obvious. Since top ( S ) is a topology, and every set in S is open with respect to top ( S ), it follows that every finite intersection of sets in S is again in top ( S ), which means that every set in V ( S ) is again open with respect to top ( S ). But then arbitrary unions of sets in V ( S ) are again open with respect to top ( S ). To prove the implication we define T = A X : V A V ( S ) such that A = [ B V A B , and we will show that (1) top ( S ) { X } T . By the definition of top ( S ) it suffices to prove the following Claim : The collection T 1 = { X } T is a topology on X , which contains all the sets in S ....
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This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

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03-top-constr - Lecture 3 3. Constructing topologies In...

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