Normal x Compact Needs Not be Normal

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 X=[0,\omega_1) and the compact space is Y=[0,\omega_1]. I would like to show that X \times Y 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 X \times Y is a subspace of Y \times Y. Thus Y \times Y 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 [0,\omega_1) and [0,\omega_1] 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 X \times Y is not normal, the Pressing Down Lemma is used. A subset S of X=[0,\omega_1) is a stationary set if S meets every closed and unbounded set in X. The following is one version of the Pressing Down Lemma (see Lemma 6.15 on p. 80 of [1] for a more general version).

    Pressing Down Lemma

      Let S be a stationary subset of [0,\omega_1). Let f:S \rightarrow \omega_1 such that for each \gamma \in S, f(\gamma)<\gamma, then for some \alpha<\omega_1, f^{-1}\lbrace{\alpha}\rbrace is a stationary subset of [0,\omega_1).

To show that X \times Y is not normal, let H and K be the following sets.

    H=\lbrace{(\alpha,\alpha):\alpha<\omega_1}\rbrace

    K=\lbrace{(\alpha,\omega_1):\alpha<\omega_1}\rbrace

These two sets are disjoint and closed in X \times Y. Let U and V be open such that H \subset U and K \subset V. It follows that U \cap V \neq \phi.

For each \alpha<\omega_1, let g(\alpha)<\omega_1 such that [g(\alpha),\alpha] \times [g(\alpha),\alpha] \subset U. Since g is a pressing down function, there is some \delta<\omega_1 and there is a stationary set S such that all points in S are mapped to \delta by g. Specifically, for each \alpha \in S, we have:

    [\delta,\alpha] \times [\delta,\alpha] \subset U

Choose \beta_0>\delta and let \beta=\beta_0+1. We have (\beta,\omega_1) \in K \subset V. Choose \gamma>\delta such that

    \lbrace{\beta}\rbrace \times [\gamma,\omega_1] \subset V

Choose \alpha \in S such that \delta < \beta<\alpha and \delta<\gamma<\alpha<\omega_1. We have the following set inclusions:

    \lbrace{\beta}\rbrace \times [\gamma,\alpha] \subset [\delta,\alpha] \times [\delta,\alpha] \subset U

    \lbrace{\beta}\rbrace \times [\gamma,\alpha] \subset \lbrace{\beta}\rbrace \times [\gamma,\omega_1] \subset V

Thus U \cap V \neq \phi. This completes the proof that the product X \times Y is not normal.

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Reference

  1. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

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Revised February 3, 2014.
\copyright \ 2014 \text{ by Dan Ma}

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