In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.
The following theorems are the main results in this post.
Let and be countably tight spaces. If one of and is compact, then is countably tight.
The product of finitely many compact countably tight spaces is countably tight.
Suppose that are countably many compact spaces such that each has at least two points. If each is a countably tight space, then the product space is countably tight.
Before proving Theorem 1 and Theorem 2, we prove the following results.
Let be a continuous and closed map from the space onto the space . Suppose that the space is countably tight and that each fiber of the map is countably tight. Then the space is countably tight.
Proof of Theorem 4
Let and where . We proceed to find a countable such that . Choose such that .
Let be the fiber of the map at the point , i.e. . By assumption, is countably tight. Call a point countably reached by if there is some countable such that . Let be the set of all points in that are countably reached by . We claim that .
Let be open such that . Because the space is regular, choose open such that and . Then . Furthermore, . Let . By the continuity of , we have . Since is countably tight, there exists some countable such that . Choose a countable such that . It follows that .
We show that that . Since , we have . Note that is a closed set since is a closed map. Thus . As a result, . Then for some . We have .
By the definition of the set , we have . Furthermore, . Note that the arbitrary open neighborhood of contains points of . This establishes the claim that .
Since is a fiber of , is countably tight by assumption. Choose some countable such that . For each , choose a countable with . Let . Note that and is countable with . This establishes the space is countably tight at .
Let be the projection map. If is a compact space, then is a closed map.
Proof of Lemma 5
Let be a closed subset of . Suppose that is not closed. Let . It follows that no point of belongs to . For each , choose open subset of such that and . The set of all is an open cover of the compact space . Then there exist finitely many that cover , say for .
Let . We have . Since is compact, we can then use the Tube Lemma which implies that there exists open such that . It follows that . Choose . Then for some , . Since , , implying that , a contradiction. Thus must be a closed set in . This completes the proof of the lemma.
Proof of Theorem 1
Let be the factor that is compact. let be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space is countably tight by assumption. Each fiber of the projection map is of the form where , which is countably tight. Then use Theorem 4 to establish that is countably tight.
Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight.
Our proof to establish that the product space is countably tight is an indirect one and makes use of two non-trivial results. We first show that is a closed subspace of a -product that is normal. It follows from another result that the second factor is countably tight. We now present all the necessary definitions and theorems.
Consider a product space where is an infinite cardinal number. Fix a point . The -product of the spaces with as the base point is the following subspace of the product space :
The definition of the space depends on the base point . The discussion here is on properties of that hold regardless of the choice of base point. If the factor spaces are indexed by a set , the notation is .
If all factors are identical, say for all , then we use the notation to denote the -product. Once useful fact is that if there are many factors and each factor has at least 2 points, then the space can be embedded as a closed subspace of the -product.
For each , let be a space with at least two points. Then contains as a closed subspace. See Exercise 3 in this previous post.
Now we discuss normality of -products. This previous post shows that if each factor is a separable metric space, then the -product is normal. It is also well known that if each factor is a metric space, the -product is normal. The following theorem handles the case where each factor is a compact space.
For each , let be a compact space. Then the -product is normal if and only if each factor is countably tight.
Theorem 7 is Theorem 7.5 in page 821 of . Theorem 7.5 in  is stated in a more general setting where each factor of the -product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.
Let be a compact space. Then the product space is normal if and only if is countably tight.
We now prove Theorem 3.
Proof of Theorem 3
Let , where for each , and that if . For each , let . By Theorem 7, each is normal. Let , which is also normal. By Theorem 6, the space of countable ordinals is a closed subspace of . Let . We have the following derivation.
Recall that . The space is defined such that for each and for each , . Furthermore, for , for each , let . Thus is a -product of compact countably tight spaces and is thus normal by Theorem 7. The space is a closed subspace of the normal space . By Theorem 8, the product space must be countably tight.
Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let .
In the above derivation, is a point of for all . When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.
Let be the projection map. If is a countably compact space and is a Frechet space, then is a closed map.
Let and be countably tight spaces. If one of and is a countably compact space and the other space is a Frechet space, then is countably tight.
Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.
- Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.