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.