In this post, we revisit a counterexample that was discussed previously in this blog. A previous post called “Normal x compact needs not be normal” shows that the Tychonoff product of two normal spaces needs not be normal even when one of the factors is compact. The example is . In this post, we show that fails even to be subnormal. Both and are spaces of ordinals. Thus they are completely normal (equivalent to hereditarily normal). The second factor is also a compact space. Yet their product is not only not normal; it is not even subnormal.
A subset of a space is a subset of (or a -set in ) if is the intersection of countably many open subsets of . A subset of a space is a subset of (or a -set in ) if is a -set in (equivalently if is the union of countably many closed subsets of ).
A space is normal if for any disjoint closed subsets and of , there exist disjoint open subsets and of such that and . A space is subnormal if for any disjoint closed subsets and of , there exist disjoint subsets and of such that and . Clearly any normal space is subnormal.
A space is pseudonormal if for any disjoint closed subsets and of (one of which is countable), there exist disjoint open subsets and of such that and . The space is pseudonormal (see this previous post). The Sorgenfrey plane is an example of a subnormal space that is not pseudonormal (see here). Thus the two weak forms of normality (pseudonormal and subnormal) are not equivalent.
The same two disjoint closed sets that prove the non-normality of are also used for proving non-subnormality. The two closed sets are:
The key tool, as in the proof for non-normality, is the Pressing Down Lemma (). The lemma has been used in a few places in this blog, especially for proving facts about (e.g. this previous post on the first uncountable ordinal). Lemma 1 below is a lemma that is derived from the Pressing Down Lemma.
Pressing Down Lemma
Let be a stationary subset of . Let be a pressing down function, i.e., satisfies: . Then there exists such that is a stationary set.
Let . Suppose that where each is an open subset of . Then for some .
Proof of Lemma 1
For each and for each where is a limit, choose such that . The function can be chosen since is open in the product . By the Pressing Down Lemma, for each , there exists and there exists a stationary set such that for all . It follows that for each . Choose such that for all . Then for each .
The product space is not subnormal.
Proof of Theorem 2
Let and be defined as above. Suppose and where each and each are open in . Without loss of generality, we can assume that , i.e., is open in for each . By Lemma 1, for some .
Choose such that is a successor ordinal. Note that . For each , there exists some such that . Choose such that for all and that . Note that . It follows that . Thus there are no disjoint sets separating and .
- Kunen, K., Set Theory, An Introduction to Independence Proofs, First Edition, North-Holland, New York, 1980.