The space is the product of many copies of the two-element set where is the first uncountable ordinal. It is a compact space by Tychonoff’s theorem. It is a normal space since every compact Hausdorff space is normal. A space is hereditarily normal if every subspace is normal. Is the space hereditarily normal? In this post, we give two proofs that it is not hereditarily normal. It then follows that any product space cannot be hereditarily normal as long as there are uncountably many factors and every factor has at least two point.

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**The connection with a theorem of Katetov**

It turns out that there is a connection with a theorem of Katetov. For any compact space, knowing hereditary normality of the first several self product spaces can reveal a great deal of information about the compact space. More specifically, for any compact space , knowing whether , and are hereditarily normal can tell us whether is metrizable. If all three are hereditarily normal, then is metrizable. If one of the three self products is not hereditarily normal, then is not metrizable. This fact is based on a theorem of Katetov (see this previous post). The space is not metrizable since it is not first countable (see Problem 1 below). Thus one of its first three self products must fail to be hereditarily normal.

These two proofs are not direct proof in the sense that a non-normal subspace is not explicitly produced. Instead the proofs use other theorem or basic but important background results. One of the two proofs (#2) uses a theorem of Katetov on hereditarily normal spaces. The other proof (#1) uses the fact that the product of uncountably many copies of a countable discrete space is not normal. We believe that these two proofs and the required basic facts are an important training ground for topology. We list out these basic facts as exercises. Anyone who wishes to fill in the gaps can do so either by studying the links provided or by consulting other sources.

The theorem of Katetov mentioned earlier provides a great exercise – for any non-metrizable compact space , determine where the hereditary normality fails. Does it fail in , or ? This previous post examines a small list of compact non-metrizable spaces. In all the examples in this list, the hereditary normality fails in or . The space can be added to this list. All the examples in this list are defined using no additional set theory axioms beyond ZFC. A natural question: does there exist an example of compact non-metrizable space such that the hereditary normality holds in and fails in ? It turns out that this was a hard problem and the answer is independent of ZFC. This previous post provides a brief discussion and has references for the problem.

All spaces under consideration are Hausdorff spaces.

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**Exercises**

**Problem 1**

Let be a compact space. Show that is normal.

**Problem 2**

For each , let be a set with cardinality . Show that .

Problem 2 holds for any infinite cardinal, not just . One reference for Problem 2 is Lemma 10.21 on page 30 of *Set Theorey, An Introduction to Independence Proofs* by Kenneth Kunen.

**Problem 3**

For each , let be a space with at least two points. Show that for every point , there does not exist a countable base at the point . In other words, the product space is not first countable at every point. It follows that product space is not metrizable.

**Problem 4**

In any space, a -set is a set that is the intersection of countably many open sets. When a singleton set is a -set, we say the point is a -point. For each , let be a space with at least two points. Show that every point in the product space is not a -point.

Note that Problem 4 implies Problem 3.

For Problem 3 and Problem 4, use the fact that there are uncountably many factors and that a basic open set in the product space is of the form and that it has only finitely many coordinates at which .

**Problem 5**

For each , let be the set of non-negative integers with the discrete topology. Show that the product space is not normal.

See here for a discussion of Problem 5.

**Problem 6**

Let . Show that has a countably infinite subspace

such that is relatively discrete. In other words, is discrete in the subspace topology of . However is not discrete in the product space since is compact.

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**Proof #1**

Let . We show that is not hereditarily normal.

Note that the product space can be written as the product of many copies of itself:

The fact (1) follows from the fact that the union of many pairwise disjoint sets, each of which has cardinality , has cardinality (see Problem 2). The space has a countably infinite subspace that is relatively discrete (see Problem 6). In other words, it has a subspace that is homemorphic to where has the discrete topology. Thus the following is homeomorphic to a subspace of .

By Problem 5, the space is not normal. Hence the compact space contains the non-normal space and is thus not hereditarily normal.

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**Proof #2**

Let . We show that is not hereditarily normal. This proof uses a theorem of Katetov, discussed in this previous post and stated below.

*Theorem 1*

If is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

- The factor is perfectly normal.
- Every countable and infinite subset of the factor is closed.

First, can be written as the product of two copies of itself:

This is because the union of two disjoints sets, each of which has cardinality , has carinality . Note that the countably infinite subset from Problem 6 is not a closed subset of . If it were, the compact space would contain an infinite set with no limit point. Thus the second condition of Theorem 1 is not satisfied. If were to be hereditarily normal, then the first condition must be satisfied, i.e. is perfectly normal (meaning that is normal and that every closed subset of it is a -set). However, Problem 4 indicates that no point in can be a point. Therefore cannot be hereditarily normal.

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**Corollary**

The product of uncountably many spaces, each one of which has at least two points, contains a homeomorphic copy of the space . Thus such a product space can never be hereditarily normal. We state this more formally below.

*Theorem 2*

Let be any uncountable cardinal. For each , let be a space with at least two points. Then is not hereditarily normal.

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