The product of a normal space with a compact space needs not be normal. For example, the product space is not normal where is the first uncountable ordinal with the order topology and is the immediate successor of (see this post). However, is normal where is the unit interval with the usual topology. The topological story here is that has countable tightness while the compact space does not. In this post, we prove the following theorem:
Let be an infinite compact space. Then the following conditions are equivalent:
- The product space is normal.
- has countable tightness, i.e., .
Theorem 1 is a special case of the theorem found in . The proof for the direction of countable tightness of implies is normal given in  relies on a theorem in another source. In this post we attempt to fill in some of the gaps. For the direction , we give a complete proof. For the direction , we essentially give the same proof as in , proving it by using a series of lemmas (stated below).
The authors in  studied the normality of where is not necessarily compact. The necessary definitions are given below. All spaces are at least Hausdorff.
Definitions and Lemmas
Let be a topological space. The tightness of , denoted by , is the least infinite cardinal number such that for any and for any , there exists a such that and . When , we say has countable tightness or is countably tight. When , we say has uncountably tightness or is uncountably tight. An handy example of a space with uncountably tightness is . This space has uncountable tightness at the point . All first countable spaces and all Frechet spaces have countable tightness. The concept of countable tightness and tightness in general are discussed in more details here.
A sequence of points of a space is said to be a free sequence if for each , . When a free sequence is indexed by the cardinal number , the free sequence is said to have length . The cardinal function is the least infinite cardinal such that if is a free sequence of length , then . The concept of tightness was introduced by Arkhangelskii and he proved that (see p. 15 of ). This fact implies the following lemma.
Let be compact. If , then there exists a free sequence of length .
A proof of Lemma 2 can be found here.
The proof of the direction also uses the following lemmas.
For any compact space , .
Let be a normal space. For every pair and of disjoint closed subsets of , and have disjoint closures in .
For Lemma 3, see 3.12.20(c) on p. 237 of . For Lemma 4, see 3.6.4 on p. 173 of .
Proof of Theorem 1
Let . Suppose that is normal. Suppose that has uncountable tightness, i.e., . By Lemma 2, there exists a free sequence . For each , let . Then the collection has the finite intersection property. Since is compact, . Let . Consider the following closed subsets of .
We claim that . Suppose that . Either or . The latter case is not possible. Note that is an open set containing . This open set cannot contain points of the form where . So the first case must hold. Since , , a contradiction. So and are disjoint closed subsets of .
Now consider , the Stone-Cech compactification of . By Lemma 3, . Let and (closures in ). We claim that . Let be an open set in with . Note that . Thus . Choose such that . We have and . On the other hand, . Thus , a contradiction. Since is normal, Lemma 4 indicates that and should have disjoint closures in . Thus has countable tightness.
Suppose . Let and be disjoint closed subsets of . The following series of claims will complete the proof:
For each , there exists an such that either or where
Proof of Claim 1
Let . The set is a copy of . It is a known fact that in , there cannot be two disjoint closed and unbounded sets. Let and . If and , they cannot be both unbounded in . Thus the claim follows if both and . Now suppose only one of and is non-empty. If the one that is non-empty is bounded, then the claim follows. Suppose the one that is non-empty is unbounded, say . Then and the claim follows.
For each , there exists an and there exists an open set with such that one and only one of the following holds:
Proof of Claim 2
Let . Let be as in Claim 1. Assume that . We want to show that there exists an open set with such that (1) holds. Suppose that for each open with , there is a and there exists such that . Let be the set of all such points . Then . Since has countable tightness, there exists countable such that . Since is countable, choose such that for all . Note that does not contain points of since . For each , the point has an open neighborhood that contains no point of . Since is compact, finitely many of these neighborhoods cover . Let these finitely many open neighborhoods be where . Let . Then and would contain a point of , say . Then for some , a contradiction. Note that is a point of . Thus there exists an open with such that (1) holds. This completes the proof of Claim 2.
For each , there exists an and there exists an open set with such that there are disjoint open subsets and of with and .
Proof of Claim 3
Let . Let and be as in Claim 2. Assume (1) in the statement of Claim 2 holds. Note that is a product of two compact spaces and is thus compact (and normal). Let and be disjoint open sets in such that and . Note that contains no points of . Then and are the desired open sets. This completes the proof of Claim 3.
To make the rest of the proof easier to see, we prove the following claim , which is a general fact that is cleaner to work with. Claim 4 describes precisely (in a topological way) what is happening at this point in the proof.
Let be a space. Let and be disjoint closed subsets of . Suppose that is a collection of open subsets of covering such that for each , only one of the following holds:
Then there exist disjoint open subsets of separating and .
Proof of Claim 4
Let and . Note that . Likewise, . Let and . Then and are disjoint open sets. Furthermore, and . This completes the proof of Claim 4.
Now back to the proof of Theorem 1. For each , let , and be as in Claim 3. Since is compact, there exists such that is a cover of . For each , let and . Note that both and are open in . To apply Claim 4, rearrange the open sets and and re-label them as . By letting , and , the open sets satisfy Claim 4. Tracing the to or and then to and , it is clear that the two conditions in Claim 4 are satisfied:
Then by Claim 4, the disjoint closed sets and can be separated by two disjoint open subsets of .
The theorem proved in  is essentially the statement that for any compact space , the product is normal if and only . Here is the first ordinal of the next cardinal that is greater than .
- Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
- Gruenhage, G., Nogura, T., Purisch, S., Normality of , Topology and its Appl., 39, 263-275, 1991.
- Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
- Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.