$T_4$-ness that is preserved by product

Question

Sorgenfrey line demonstrates how normality can be not preserved when "squared."

Is there an example for a normal space $X$ for each of?:

• $X^2$ is normal, but $X^3$ is not

• $X^2$ and $X^3$ are normal, but $X^4$ is not

• $X^k$ is normal for $2 \leq k < n$, but $X^n$ is not

• $X^n$ is normal for $n \in \mathbb N$, but $X^\omega$ is not

• $X^n$ is normal for $n \in \mathbb N$, but $X^\omega$ is not, in box topology

Answer 1

For the first four questions, the answer is "yes", and I believe was first established in the paper

• Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.

Therein the following theorem is proved:

Theorem 1.1. For every $k$ and $m$ such that $1 \leq k \leq m \leq \omega$ there exists a separable and first-countable space $X = X ( k , m)$ such that

1. $X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$;
2. $X^n$ is normal (collectionwise normal) if and only if $n < m$.

The answer to the last question is also "yes", and an example was essentially established in

• van Douwen, Eric K., Another nonnormal box product, General Topology Appl. 7, 71-76 (1977). ZBL0341.54008.

Theorem B. $\Box^\omega ( 2^{\omega_2} )$ is not normal.

(I.e., the box product of $\omega$ copies of the Tychonoff product $2^{\omega_2} = \{ 0,1 \}^{\omega_2}$ is not normal.) Clearly $2^{\omega_2}$ is compact Hausdorff, hence normal, and each finite Tychonoff product $( 2^{\omega_2} )^n$ is homeomorphic to $2^{\omega_2}$.