Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii (see Problem I.5.25 in ). One partial positive answer is a theorem attributed to Corson: if is a normal dense subspace of a product of separable spaces such that is normal, then is collectionwise normal. Another partial positive answer: assuming , any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). Another partial positive answer to Arkhangelskii’s question is the theorem due to Reznichenko: If , which is a dense subspace of the product space , is normal, then it is collectionwise normal (see Theorem I.5.12 in ). In this post, we highlight another partial positive answer to the question posted in . Specifically, we prove the following theorem:
Let be a product space where each factor is a separable metric space. Let be a dense subspace of . Then if is normal, then is collectionwise normal.
Since any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post), it suffices to prove the following theorem:
Let be a product space where each factor is a separable metric space. Let be a dense subspace of . Then if is normal, then every closed and discrete subspace of is countable, i.e., has countable extent.
Arkhangelskii’s question was studied by the author of  and . Theorem 1 as presented in this post is essentially the Theorem 1 found in . The proof given in  is a beautiful proof. The proof in this post is modeled on the proof in  with the exception that all the crucial details are filled in. Theorem 1a (as stated above) is used in  to show that the function space contains no dense normal subspace.
It is natural to wonder if Theorem 1 can be generalized to product space of many separable metric factors where is an arbitrary uncountable cardinal. The work of  shows that the question at the beginning of this post cannot be answered positively in ZFC. Recall the above mentioned result that assuming , any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). A theorem in  implies that assuming , for any separable metric space with at least 2 points, the product of continuum many copies of contains a normal dense subspace that is not collectionwise normal. A side note: for this normal subspace , is necessarily not normal (according to Corson’s theorem). Thus  and  collectively show that Arkhangelskii’s question stated here at the beginning of the post is answered positively (in ZFC) among product spaces of many separable metric factors and that outside of the case, it is impossible to answer the question positively in ZFC.
Proving Theorem 1a
We use the following lemma. For a proof of this lemma, see the proof for Lemma 1 in this previous post.
Let be a product of separable metrizable spaces. Let be a dense subspace of . Then the following conditions are equivalent.
- is normal.
- For any pair of disjoint closed subsets and of , there exists a countable such that .
- For any pair of disjoint closed subsets and of , there exists a countable such that and are separated in , meaning that .
For any , let be the natural projection from the product space into the subproduct space .
Proof of Theorem 1a
Let be a dense subspace of the product space where each factor has a countable base. Suppose that is an uncountable closed and discrete subset of . We then construct a pair of disjoint closed subsets and of such that for all countable , and are not separated, specifically . Here the closure is taken in the space . By Lemma 2, the dense subspace of is not normal.
For each , let be a countable base for the space . The standard basic open sets in the product space are of the form such that
- each is an open subset of ,
- if , then ,
- for all but finitely many .
We use to denote the finite set of such that . Technically we should be working with standard basic open subsets of , i.e., sets of the form where is a standard basic open set as described above. Since is dense in the product space, every standard open set contains points of . Thus we can simply work with standard basic open sets in the product space as long as we are working with points of in the construction.
Let be the collection of all standard basic open sets as described above. Since there are only many factors in the product space, . Recall that is an uncountable closed and discrete subset of . Let be the following:
Claim 1. .
First we show that . Let be countable. Consider these two cases: Case 1. is an uncountable subset of ; Case 2. is countable.
Suppose Case 1 is true. Since is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point such that every open neighborhood of (open in ) contains uncountably many points of . Thus every standard basic open set , with , contains uncountably many points of . Suppose Case 2 is true. There exists one point such that for uncountably many . Then in either case, every standard basic open set , with and , contains uncountably many points of . Any one such is a member of .
We can partition the index set into many disjoint countable sets . Then for each such , obtain a in either Case 1 or Case 2. Since , all such open sets are distinct. Thus Claim 1 is established.
There exists an uncountable such that for each , and .
Enumerate . Choose with . Suppose that for all , two points are chosen such that , and such that and where . Then choose with such that and and where .
Let and let . Note that . Based on the inductive process that is used to obtain and , it is clear that satisfies Claim 2.
For each countable , the sets and are not separated in the space .
Let be countable. Consider the two cases: Case 1. is uncountable; Case 2. is countable. Suppose Case 1 is true. Since is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point such that every open neighborhood of (open in ) contains uncountably many points of . Choose such that . Then the following statement holds:
- For every basic open set with such that , the open set contains uncountably many points of .
Suppose Case 2 is true. There exists some such that for uncountably many . Choose such that . Then statement 1 also holds.
In either case, there exists such that statement 1 holds. The open sets described in statement 1 are members of . By Claim 2, the open sets described in statement 1 also contain points of . Since the open sets described in statement 1 have supports , the following statement holds:
- For every basic open set with , the open set contains points of .
Statement 2 indicates that . Thus . The closure here can be taken in either or (to apply Lemma 2, we only need the latter). Thus Claim 3 is established.
Claim 3 is the negation of condition 3 of Lemma 2. Therefore is not normal.
The proof of Theorem 1a, though a proof in ZFC only, clearly relies on the fact that the product space is a product of many factors. For example, in the inductive step in the proof of Claim 2, it is always possible to pick a pair of points not chosen previously. This is because the previously chosen points form a countable set and each open set in contains many points of the closed and discrete set . With the “ versus ” situation, at each step, there are always points not previously chosen. When more than many factors are involved, there may be no such guarantee in the inductive process.
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- Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
- Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
- Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.