This post is a basic discussion of Bing’s Example G. The original post was published on 10/27/2009 and is now replaced by a new post. The following link will take you there. Thank you.
Let be the discrete space the natural numbers. It is known that is not normal (see a proof here). It turns out that the product of uncountably many non-compact metric spaces is never normal. In 1948, A. H. Stone proved that the uncountable product of metric spaces is normal if and only if all but countably many factors are compact. Thus product of uncountably many copies of is never normal.
Theorem. Let be a family of metrizable spaces. The following conditions are equivalent.
(1) is paracompact.
(2) is normal.
(3) All but countably many are compact.
Proof. is obvious.
. Suppose is normal. I Claim that all but countably many factors are countably compact. When this claim is established, (3) is established. Note that if a paracompact space is countably compact, it is compact. Suppose that there are uncountably many that are not countably compact where . Then each such would contain a closed copy of . Thus the product space would contains as a closed subspace. This is a contradiction since any closed subspace of a normal space is normal. Thus all but countably many factors are countably compact.
. Suppose all but countably many are compact. Then where is the product of all the compact factors and is the product of the countably many non-compact factors. Note that is also a metrizable space. The product of a compact space and a paracompact space is paracompact (see a proof here). Thus (1) is established.
Corollary. Let be a family of separable metrizable spaces. The following conditions are equivalent.
(1) is paracompact.
(2) is Lindelof.
(3) is normal.
(4) All but countably many are compact.
The only thing I want to mention about the corollary is that being a product of separable spaces, the product space has the countable chain condition (ccc). In any space with the ccc, paracompactness implies the Lindelof property (see a proof here).
[Stone] Stone, A. H.,  Paracompact and Product Spaces, Bull. Amer. Math. Soc., 54, 977-982.
The theorem indicated by the title is a basic result and is an interesting theorem. It is known that “compact x normal” needs not be normal (see an example here). But for a compact factor and a paracompact factor, the product is not only normal, it is paracompact. I would like to write down a proof. This theorem will be used in subsequent posts. If the factor is Lindelof instead, it can be shown that is Lindelof.
Let be a compact space and let be a paracompact space. Let be an open cover of . For each , let be a finite cover of .
Claim. Fix . There is an open set such that . For each , let be open and let be open such that , , and . Since is compact, choose finitely many open sets that cover , say . Let . It is clear that .
The collection is an open cover of . Then it has a locally finite open refinement . For each , choose such that . Consider where . Let be the collection of all such open sets .
There are three things to check here. One is that is an open cover of . Fix . Then for some . Note that for some . This means that the point for some in .
The second to check is that is a refinement of the original open cover . This is clear since every set in is chosen to be a subset of some set in .
The third is that is a locally finite collection. To see this, let . There is an open such that and can only meet finitely many . For each such , is associated with finitely many sets in . This means is an open set containing that can meet only finitely many sets in . This completes the proof that is paracompact.
If the factor Y is Lindelof, then we can modify the proof to show that is Lindelof. In the step above where is obtained, we get a countable subcover of . Each member of this countable subcover is associated with a finite . Thus we can obtain a countable subcover of .
In a previous post, I showed that the uncountable product of the space of the integers is not normal. Consequently, the uncountable product of the unit interval is never hereditarily normal since it contains a copy of the uncountable product of the integers. It turns out that, as the title of this post suggests, hereditarily normality cannot happen in uncountable product of spaces. Specifically the product of uncountably many spaces, each of which has at least two points, is never hereditarily normal.
Let such that and each has at least two points. The proof is quite simple, a matter of rearranging the factors so that the product space contains a copy of .
For each , let be a two-point subspace. Break up into many disjoint subsets , where . For each , let , which is homeomorphic to the middle third Cantor set in the unit interval. Thus, .
Each Cantor set contains a subset that is homeomorphic to the discrete space of the natural numbers . Thus .
The Tychonoff plank is the product space , that is, the Tychonoff Plank is the product of the successor of the first uncountable ordinal and the successor of the first infinite ordinal. The deleted Tychonoff plank is , i.e., the Tychonoff plank minus the corner point . This is another famous example of topological spaces that often appear in first year topology courses. My goal here is to present two facts. The deleted Tychonoff plank is not normal, thus showing that the Tychonoff plank not hereditarily normal. The second point is that for the deleted Tychonoff plank, the one-point compactification and the Stone-Cech compactification conincide. This is due to the fact that any continuous real-valued function defined on the deleted Tychonoff plank can be extended to a continuous function on the product space (see Theorem 19.12 in [Willard]).
First we show the deleted Tychonoff plank is not normal. Let and . These are two disjoint closed sets in . Let and where and are open in . I will show .
First define a pressing down function on . For each limit ordinal , there is an integer and there is a countable ordinal such that . By the pressing down lemma, there is and there is a stationary set such that all points in are mapped to by . Choose an uncountable such that for each , for some . Consider .
Claim. where is the closure operator. Fix . Fix an open set containing , say where . Choose such that . Based on how is obtained, . It follows that and . So any open set containing contains points of . Thus the claim is established.
Since , would contain points of . Thus the deleted Tychonoff plank is not normal.
To established the second point, let a continuous function. Then can be extended to include the corner point. Consider the restriction of to the following horizontal segments of the deleted Tychonoff plank.
for each .
Any continuous function is eventually constant on these horizontal segments. For the proof, see this post. So there is such that for all . For each , there is such that for all . Choose some for all .
It is clear that the sequence converges to . So by defining , is still a continuous function.
[Willard] Willard, S.,  General Topology, Addison-Wesley Publishing Company, Inc.
This is to present a non-normal cross product where one factor is a normal space and the other factor is a compact space. The normal space is and the compact space is . I would like to show that is not normal. For a basic discussion on these ordinal spaces with the order topology, see this previous post.
In addition to demonstrating an example meeting the criteria set forth in the title, I would like to look at this through two angles. Because these spaces have the ordered topology, they are hereditarily normal (completely normal). Note that is a subspace of . Thus fails to be hereditarily normal, demonstrating that hereditarily normal x hereditarily normal needs not be hereditarily normal. Another example of this is the double arrow space. In this previous post, I showed that the square of the double arrow space is not hereditarily normal. The double arrow space is perfectly normal and is thus a stronger example in this regard since it shows the square of a perfectly normal space needs not be hereditarily normal. But the uncountable ordinal spaces of and are important counterexamples as well as building blocks for other counterexamples. So for the record, I present them here.
For the second angle, these two spaces show that hereditarily normality is not “nice” enough to prevent non-normal product.
In proving is not normal, the Pressing Down Lemma is used. A subset of is a stationary set if meets every closed and unbounded set in . The following is one version of the Pressing Down Lemma (see Lemma 6.15 on p. 80 of  for a more general version).
Pressing Down Lemma
Let be a stationary subset of . Let such that for each , , then for some , is a stationary subset of .
To show that is not normal, let and be the following sets.
These two sets are disjoint and closed in . Let and be open such that and . It follows that .
For each , let such that . Since is a pressing down function, there is some and there is a stationary set such that all points in are mapped to by . Specifically, for each , we have:
Choose and let . We have . Choose such that
Choose such that and . We have the following set inclusions:
Thus . This completes the proof that the product is not normal.
- Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.
Revised February 3, 2014.
It is a well known fact in general topology that in the class of spaces with the countable chain condition (CCC), paracompactness equals the Lindelof property. Any regular Lindelof space is paracompact. This post gives a proof that any paracompact space with the CCC is a Lindelof space.
A space has the countable chain condition (CCC) if every pairwise disjoint family of open sets in the space is countable. Clearly any separable space has the CCC. Other examples of spaces with the CCC are product of separable spaces and function spaces where is any completely regular space.
CCC + Paracompact implies Lindelof
Suppose that the space is a paracompact with the CCC. Let be an open cover of . By the paracompactness of , there is , which is a locally finite open refinement of .
For each , choose such that and choose , an open subset of such that and meets only finitely many elements of . For each , let . Let . One observation about is that it is an open cover of . Another observation about is that each element of meets only finitely many elements of .
For any , the finite set is said to be a chain from to if , and for all . For each , define and as follows:
Observe that each is a countable subcollection of . For , if , then there is a chain from to . It follows that and . Let be the family of all distinct where . By the CCC of the space , must be countable.
Note that is also an open cover of . As observed previously, for each , is countable. For each , for some . Thus each is matched with countably many . Since is countable, there are countably many that associate with the elements of . Since is a cover of , the associated countably many must form an open cover of too. It follows that countably many elements of the original cover form a cover of . This concludes the proof that is Lindelof.
Dan Ma topology
Daniel Ma topology
Dan Ma math
Daniel Ma mathematics
Revised June 15, 2019