I am looking at some topological dimension theory for product spaces, and in trying to construct a certain type of counterexample it's become relevant to consider the question in the title above. I am interested in finding a normal space $X$ whose products with itself is eventually non-normal, but not immediately.

It's not actually important for my application that it happens in three steps as opposed to more. An alternative question would be: Is there a normal space $X$ with $X \times X = Y$ normal, but $Y \times Y$ is not normal?

The original problem is here:


Thanks for any help!

As mentioned in a comment below, if we assume that $X$ is a compact Hausdorff space and that $X \times X \times X$ is completely normal, then $X$ is metrizable. Thus it stands to reason that a compact counterexample may be harder (if not impossible) to construct. The author in the linked paper wonders aloud if the complete normality of $X \times X$ is sufficient for the metrizability of $X$, so it may also be advisable to avoid cases where $X \times X$ is completely normal.


A construction can be found in (or weaned from)

wherein the following remarkable result is proved:

Theorem 1. For every $k$ and $m$ such that $1 \leq k \leq m \leq \omega$ there exists a separable and first coutnable 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 $n < m$.

In particular, we can construct a (normal) space such that the failure of normality of its powers happens first at any prescribed finite power.

  • $\begingroup$ Wow, what a result! $\endgroup$ – John Samples Nov 21 '18 at 10:06

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