We continue on our quest to prove the Tychonoff Theorem: an arbitrary product of compact spaces is compact. We just show that this is true for the FINITE product of compact spaces.
It is our goal to do this by using elementary tools and avoiding things like nets (for example, Willard uses ultranets)
We will basically adding background and commentary to David Wright’s excellent 1994 paper which appeared in the Proceedings of the American Mathematical Society. We will use a bit of cardinal arithmetic and facts about ordinals at times.
Yes, we do need some background, but the background we are providing is necessary for the understanding of any mathematics that uses topology anyway.
Conditions which are equivalent to compactness
1. If in the usual topology, then is compact if is closed and bounded.
Proof: Let be compact. Then is closed because the usual topology for is Hausdorff. is bounded as well, as if it weren’t, the if latex X $ by . This open cover has no finite subcover as is unbounded.
Now let be closed and bounded. Then for some real , which is compact by our “junior” Tychonoff Theorem. So is a closed subset of a compact set and therefore compact.
2. I’ll call this the excluded point condition: Let , where each is open. We say that lacks a finite subcover if there is no finite subcollection of the that covers U. A topological space is said to have the excluded point condition if any subset which has an open cover which lacks a finite subcover must exclude at least one point of ; that is, any set which has an open cover with no finite subcover must be a proper subset of .
Example of an open cover which lacks a finite subcover: consider as a subset of in the usual topology, and let ; here are the excluded points from this open cover.
Theorem: a space is compact if and only if it has the excluded point condition for open covers.
Proof. If is compact then any open cover of has a finite subcover, hence any subset of which has an open cover which lacks a finite subcover cannot be all of .
Now assume that has the excluded point condition. Let be any open cover of . cannot lack a finite subcover as any subset which has an open cover lacking a finite subcover must exclude a point of .
3. The finite intersection property condition: let be any collection of closed sets. We say that has the finite intersection property if the following holds: if the intersection of any finite subcollection of elements of is non-empty.
Example: in , has the finite intersection property. On the other hand, does not have this property as there are finite subcollections of this set that have an empty intersection.
Theorem: is compact if and only if the following holds: if is a collection of closed sets with the finite intersection property, then an arbitrary intersection of elements of is non-empty.
Proof: Let be compact. Let be an infinte collection of closed sets with the finite intersection property. This means that no finite collection of the complements of these sets can cover Then is an open cover of a subset of . Since no finite subcollection of these closed set complements (open sets) can cover all of , then is non-empty and therefore so is
Now let the finite intersection property hold for . Let be any open cover for . This means that is empty. Hence the collection of closed sets cannot have the finite intersection property which means that there is some finite subcollection where is empty which means covers .
4. Limit point compactness: part I. Theorem: a space is compact if and only if every infinite subset has a limit point.
Note: we can actually prove a bit more; that will be in part II. This is a “junior theorem” which can lead the beginner to understanding the “varsity theorem”.
Proof. Let be compact and let be an infinite subset of . Consider: where is some open set containing . If has no limit point we can assume that the are chosen so that each is finite for each . Now a finite subcollection of the covers..say and . This is impossible as each is finite but is infinite.
Now assume that is limit point compact in that every infinite subset has a limit point. Let be an open cover which has no finite subcover. We assume that this open cover is efficient in that for each ; that is, any in the open cover contains at least one point not contained in the union of the other open cover sets. Then the set is an infinite set with no limit point.
5. Let be a set with cardinality . We say that is a perfect limit point of if for all open sets containing , . Example: has every point as a perfect limit point (usual topology) as has the cardinality of the real numbers and if is open in and contains any point of then has the cardinality of the real numbers.
Now the stronger theorem is this: a topological space is compact if and only if every infinite subset has a perfect limit point.
Proof. First, assume that is compact. Let be an infinite subset with cardinality . Cover by open sets where . Suppose that for all . Now this open cover of has a finite subcover and so we have and so where each . This is impossible because is an infinite cardinal (a limit cardinal) and it is impossible to reach a limit cardinal by a finite sum of strictly smaller cardinals.
If you are new to this and are a bit confused, start by assuming that is, say, the first countably infinite cardinal. ALL lesser cardinals are finite cardinals, and it is impossible for a finite sum of finite cardinals to add up to any infinite cardinal. Then, imagine being the first uncountable cardinal. One can not reach any uncountable cardinal by the finite sum of countable (or smaller) cardinals (the finite sum of countable cardinals is still countable). That is more or less what is going on here.
Now, suppose that every infinite set has a perfect limit point. Let be an open cover which has no finite subcover. We can assume that is the index of smallest cardinality for which this is true and that the cover is efficient: that is, the open subcover is built by adding open sets which contain at least one point not contained by the previously added open sets. Also we put a well ordering on where the cardinality of . If this confuses you a bit, think of a countable index set where the cardinality of the previous indices are finite, or of an uncountable index set where the smaller cardinals are all countable.
So, for each let . Now let and note by design.
Now if , there is some where but as all have smaller cardinality than Therefore has no perfect limit point.
Again, the person new to topology can run through this with first being the countable ordinal (and every previous ordinal having finite cardinality) or being the first uncountable ordinal with every previous ordinal having at most countable cardinality.
We now have the background to give a simple proof of the full strength Tychonoff Theorem, which we will do in the next post.