Let be the first uncountable ordinal. The topology on we are interested in is the ordered topology, the topology induced by the well ordering. The space is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of . We also make comments on normality in the powers of a countably compact non-compact space.
Let be the first infinite ordinal. It is well known that , the product space of many copies of , is not normal (a proof can be found in this earlier post). This means that any product space , with uncountably many factors, is not normal as long as each factor contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space , the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of ). In other words, the interesting case is that each factor is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in where is a countably non-compact space. In this post we start with the space of the countable ordinals. We examine power as well as the countable power . The former is not normal while the latter is normal. The proof that is normal is an application of the normality of -product of the real line.
The uncountable product
The product space is not normal.
Theorem 1 follows from Theorem 2 below. For any space , a collection of subsets of is said to have the finite intersection property if for any finite , the intersection . Such a collection is called an f.i.p collection for short. It is well known that a space is compact if and only collection of closed subsets of satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.
In the space , there is an f.i.p. collection of cardinality using its linear order. For each , let . Let . It is a collection of closed subsets of . It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space with an f.i.p. collection of cardinality that has empty intersection, the product space is not normal.
Let be a countably compact space. Suppose that there exists a collection of closed subsets of such that has the finite intersection property and that has empty intersection. Then the product space is not normal.
Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space , we mean a set of the form such that each is an open subset of and that for all but finitely many . Given a standard basic open set , the notation refers to the finite set of for which . For any set , the notation refers to the projection map from to the subproduct . Each element can be considered a function . By , we mean .
For each , let be the constant function whose constant value is . Consider the following subspaces of .
Both and are closed subsets of the product space . Because the collection has empty intersection, . We show that and cannot be separated by disjoint open sets. To this end, let and be open subsets of such that and .
Let . Choose a standard basic open set such that . Let . Since is the support of , it follows that . Since has the finite intersection property, there exists .
Define such that for all and for all . Choose a standard basic open set such that . Let . It is possible to ensure that by making more factors of different from . We have . Since has the finite intersection property, there exists .
Now choose a point such that for all and for all . Continue on with this inductive process. When the inductive process is completed, we have the following sequences:
- a sequence of point of ,
- a sequence of finite subsets of ,
- a sequence of points of
such that for all , for all and . Let . Either is finite or is infinite. Let’s examine the two cases.
Suppose that is infinite. Since is countably compact, has a limit point . That means that every open set containing contains some . For each , define such that
From the induction step, we have for all . Let , the constant function whose constant value is . It follows that is a limit of . This means that . Since , .
Suppose that is finite. Then there is some such that for all . For each , define such that
As in Case 1, we have for all . Let , the constant function whose constant value is . It follows that for all . Thus .
Both cases show that . This completes the proof the product space is not normal.
The countable product
The product space is normal.
Proof of Theorem 3
The proof here actually proves more than normality. It shows that is collectionwise normal, which is stronger than normality. The proof makes use of the -product of many copies of , which is the following subspace of the product space .
It is well known that is collectionwise normal (see this earlier post). We show that is a closed subspace of where . Thus is collectionwise normal. This is established in the following claims.
We show that the space is embedded as a closed subspace of .
For each , define such that for all and for all . Let . We show that is a closed subset of and is homeomorphic to according to the mapping .
First, we show is closed by showing that is open. Let . We show that there is an open set containing that contains no points of .
Suppose that for some , . Consider the open set where except that . Then and .
So we can assume that for all , . There must be some such that . Otherwise, . Since , there must be some such that . Now choose the open interval and the open interval . Consider the open set such that except for and . Then and . We have just established that is closed in .
Consider the mapping . Based on how it is defined, it is straightforward to show that it is a homeomorphism between and .
The -product has the interesting property it is homeomorphic to its countable power, i.e.
Because each element of is nonzero only at countably many coordinates, concatenating countably many elements of produces an element of . Thus Claim 2 can be easily verified. With above claims, we can see that
Thus is a closed subspace of . Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that is normal.
The normality in the powers of
We have established that is not normal. Hence any higher uncountable power of is not normal. We have also established that , the countable power of is normal (in fact collectionwise normal). Hence any finite power of is normal. However is not hereditarily normal. One of the exercises below is to show that is not hereditarily normal.
Theorem 2 can be generalized as follows:
Let be a countably compact space has an f.i.p. collection of closed sets such that . Then is not normal where .
The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.
Let be a countably compact space that is not compact. Then there exists a cardinal number such that is not normal and is normal for all cardinal number .
By the non-compactness of , there exists an f.i.p. collection of closed subsets of such that . Let be the least cardinality of such an f.i.p. collection. By Theorem 4, that is not normal. Because is least, any smaller power of must be normal.
Let be a space that is not countably compact. Then is not normal for any cardinal number .
Since the space in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, is not normal. Then is a closed subspace of .
Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space , is not normal for some cardinal number . The from either Theorem 5 or Theorem 6 is at least . Interestingly for some spaces, the can be much smaller. For example, for the Sorgenfrey line, . For some spaces (e.g. the Michael line), .
Theorems 4, 5 and 6 are related to a theorem that is due to Noble.
Theorem 7 (Noble)
If each power of a space is normal, then is compact.
A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of is just another way of discussing Theorem 7. According to Theorem 7, if is not compact, some power of is not normal.
The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in for more examples of countably compact non-compact . One particular interesting example would be a countably compact non-compact such that the least power for non-normality in is more than . A possible candidate could be the second uncountable ordinal . By Theorem 2, is not normal. The issue is whether the power and countable power are normal.
Show that is not hereditarily normal.
Show that the mapping in Claim 3 in the proof of Theorem 3 is a homeomorphism.
The proof of Theorem 3 shows that the space is a closed subspace of the -product of the real line. Show that can be embedded in the -product of arbitrary spaces.
For each , let be a space with at least two points. Let . The -product of the spaces is the following subspace of the product space .
The point is the center of the -product. Show that the space contains as a closed subspace.
Find a direct proof of Theorem 3, that is normal.