Wednesday, September 27, 2006

Topological Spaces and Open Sets

In the previous post, when I talked about continuity, I based this on the notion of a point being near to a set. We said that a point p was near to a set A if every neighborhood of the point intersected A, and we said that in our case neighborhoods on the real line were open intervals of the form (a,b), where this is all points x such that a < x < b. We then said that a function f was continuous if for any set A and point p near to it, f(p) is near to f(A)={f(a) a Î A}.

Well, this should come as no surprise to mathematicians looking at this blog, but I was lying then. Neighborhoods can be much more complicated than just open intervals, and in fact this complexity is what makes topology interesting and difficult. We can either approach this by first studying metric spaces and deducing topology from there, as Mark did on Good Math, Bad Math,or start directly from the abstract notion of topological spaces, which is what i'll do.


Basically, a topological space X is just a set of elements, which we shall call points from now on,with a specified family U={Ua}a ÎJ of subsets of X, called the open sets of X (the J is an indexing set, as explained in the next paragraph).


Before we continue, we need a small side trip. Topologies are usually interesting only when the number of points and the number of open sets are uncountable. To be able to work with those sets, we can't use any countable set of indices (such as the naturals), and therefore we use an indexing set J which is simply some set of unspecified cardinality.


The open sets U of X must fulfill some axioms:

  1. X and the empty set Æ must be open.
  2. Any union of open sets must be open. Formally, this means that if we have a subset J¢ Í J of the indexing set, then Èa ÎJ¢ Ua Î U.
  3. Any finite intersection of open sets must be open. Formally, this means that if we have a subset {j1,¼,jn} Í J of the indexing set, then Çi=1nUji Î U.
We can now formulate a formal condition for a set being open. A set U Í X is open if and only if for each point p ÎU there is an open set Up such that p Î UpÍ U. Informally, a set U is open if and only if each of its points has a "small enough" open set around it, that is an open set which is a subset of U.


We can prove this easily, in what looks like cheating. If U is open, then for every point p we can take Up=U, and thus have the necessary open set.


For the other direction, we notice that we can write U=Èp Î U Up. At first glance, this looks very weird. p ÎUp for every p Î U, and so we definitely have U ÌÈp Î UUp. However, it seems impossible that ÈpÎ UUp ÌU, since it would seem that taking all the points and the sets around them would give us a set bigger then the original set U. Here the "small enough" demand comes into play. Since each of the sets Up sits completely inside U, their union must also sit completely inside U. To finish off, we remember that each of the Upare open by assumption, and since an arbitrary union of open sets is open, this means that U=ÈpÎ UUp is open.


This little tautology gives us the connection between the definitions of nearness and open sets. We can reformulate the notion of a point being near to a set precisely. A point p is near to a set A if for every open set U such that p ÎU, we have UÇA being nonempty.


This means that a set U is open if and only if non of its points are near to its complement X-U. For if p Î U,and U is open, there is an open set Up such that p Î Upand Up sits completely inside U, that is Up Ì U. Thus, of course,Up does not intersect the complement of U. Conversely, if none of the points in U are near to X-U, this means that every point p Î U has a"small enough" open set around it which does not intersect X-U, and we can again show that U is the union of all these "small enough" open sets and thus it is open.


This fact makes open sets the natural setting for questions of continuity, or any other concept formulated by nearness.


As a little exercise, I write down this claim, which is usually given as the definition of continuity, and the proof will come next time.


Proposition 1 A function f:X® Y between two topological spaces is continuous if and only for any set U Ì Y which is open, the set f-1(U)={xÎ X f(x) ÎU} is open in X.


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