A note about sigma-product of compact spaces

This is a basic discussion on \Sigma-products of compact spaces. Let \displaystyle Y=\Pi_{\alpha \in S} X_\alpha be a product space and let p \in Y. The \Sigma-product of the spaces \lbrace{X_\alpha}\rbrace_{\alpha \in S} about the fixed point p is the following subspace of the product space Y:

\displaystyle \Sigma_{\alpha \in S} X_\alpha=\lbrace{y \in Y:\lvert \lbrace{\alpha \in S:y(\alpha) \neq p(\alpha)}\rbrace \lvert \thinspace \leq \omega}\rbrace.

The \Sigma-products had been introduced in this blog. In a previous post, it was shown that the \Sigma-product of uncountably many separable spaces is an example of a space that has the countable chain condition (ccc) that is not separable. In this post, we discuss the \Sigma-product of compact spaces. First, the \Sigma-product of uncountably many spaces can never be paracompact because it always contains a closed copy of the first uncountable ordinal \omega_1. Second, the \Sigma-product of compact spaces may not be normal. Our example is that \displaystyle \Sigma_{\alpha \in S} X_\alpha is not normal where each X_\alpha=\omega_1+1=[0,\omega_1]. While it may be too much to expect that the Tychonoff theorem (the product space is compact if and only if each factor is compact) would have a counterpart for \Sigma-products, we do have a proof that the \Sigma-product of compact spaces is countably compact. This provides another example of a countably compact space that is non-compact. It is known that the \Sigma-product of metric spaces is normal ([1] and [2]).

When the index set S is countable, the \Sigma-product is the entire product space \displaystyle Y=\Pi_{\alpha \in S} X_\alpha. When all but countably many X_\alpha are one-point sets, the \Sigma-product can also be treated as just a product space of countably many factors. So we like to avoid these two situations by assuming that there are uncountably many spaces X_\alpha and each X_\alpha has at least two points. When this is the case, the \Sigma-product is a proper subspace of the associated product space. Such \Sigma-products are called proper.

For each \beta \in S, define \pi_\beta:Y \rightarrow X_\beta by \pi_\beta(f)=f(\beta) for each \text{f} in the product space \Pi X_\alpha. The map \pi_\beta maps each point in the product space to its \beta^{th} coordinate and is called the projection map of \Pi X_\alpha on X_\beta. Open sets of the form \bigcap_{\alpha \in F} \pi^{-1}_\alpha(V_\alpha), where F \subset S is finite and for each \alpha \in F, V_\alpha is open in X_\alpha, form a base in the product topology.

Observation. Given \displaystyle X=\Sigma_{\alpha \in S}X_\alpha and T \subset S, it is easy to verify that \displaystyle X=\Sigma_{\alpha \in S}X_\alpha is homeomorphic to \displaystyle \Sigma_{\alpha \in T}X_\alpha \times \Sigma_{\alpha \in S-T}X_\alpha. We will make use of this observation in the rest of this note.

Lemma. Let \lbrace{X_\alpha}\rbrace_{\alpha<\omega_1} be a family of spaces each of which has at least two points. Let p \in \Pi_{\alpha<\omega_1}X_\alpha. Then \displaystyle X=\Sigma_{\alpha<\omega_1}X_\alpha (about the fixed point p) contains a closed copy of the first uncountable ordinal \omega_1. Thus any proper \Sigma-product can never be paracompact.

Proof. For each \alpha<\omega_1, choose a point t_\alpha \in X_\alpha such that t_\alpha \neq p(\alpha). Consider the point q \in \Pi_{\alpha<\omega_1}X_\alpha such that q(\alpha)=t_\alpha for each \alpha<\omega_1. Since each X_\alpha is Hausdorff, choose disjoint open sets U_\alpha,V_\alpha \subset X_\alpha containing q(\alpha) and p(\alpha) respectively.

We can think of elements of X as all the functions f such that f(\alpha) \in X_\alpha for each \alpha<\omega_1 and f(\alpha)=p(\alpha) for all but countably many \alpha. For each \beta<\omega_1, consider the function f_\beta such that f_\beta(\gamma)=q(\gamma) for all \gamma<\beta and f_\beta(\gamma)=p(\gamma) for all \gamma \ge \beta. In other words, f_\beta agrees with \text{q} on the initial segment [0,\beta) and f_\beta agrees with p on the final segment [\beta,\omega_1). Then W=\lbrace{f_\beta:\beta<\omega_1}\rbrace is a closed copy of \omega_1.

Claim 1. The set W is closed in X.

We show X-W is open. Let y \in X-W. If for some \alpha \in \omega_1, y(\alpha) \in X_\alpha-\lbrace{p(\alpha),q(\alpha)}\rbrace, choose open O \subset X_\alpha such that y(\alpha) \in O and O does not contain both p(\alpha),q(\alpha). Then y \in \pi^{-1}_\alpha(O) and \pi^{-1}_\alpha(O) \cap W =\phi. Now we can assume that for each \alpha \in \omega_1, we have y(\alpha) \in \lbrace{p(\alpha),q(\alpha)}\rbrace. Since y \notin W, there are \alpha<\beta<\omega_1 such that y(\alpha)=p(\alpha) and y(\beta)=q(\beta). Then y \in \pi^{-1}_\alpha(V_\alpha) \cap \pi^{-1}_\beta(U_\beta)=B and B \cap W=\phi.

Claim 2. The mapping \alpha \rightarrow f_\alpha is a homeomorphism between \omega_1 and W.

First show that the mapping is continuous.  Let \bigcap_{\alpha \in F} \pi^{-1}_\alpha(O_\alpha) be an open set containing f_\beta where F \subset \omega_1 is finite and O_\alpha \subset X_\alpha is open for each \alpha \in F. Let V=(\gamma,\beta] be an open interval in the order topology of \omega_1 such that V misses F-\lbrace{\beta}\rbrace. For each \delta \in V, it is clear that f_\delta \in \bigcap_{\alpha \in F} \pi^{-1}_\alpha(O_\alpha). Now, we show that the inverse is continuous. Consider the open interval (\gamma,\beta]. Note that f_\beta \in \pi^{-1}_\gamma(U_\gamma) \cap \pi^{-1}_\beta(V_\beta)=O. For each f_\delta \in O, we have \delta \in (\gamma,\beta].

Example. The space \displaystyle Z=\Sigma_{\alpha<\omega_1}Z_\alpha where each Z_\alpha=\omega_1+1=[0,\omega_1] is not normal.

Let p \in (\omega_1+1)^{\omega_1}. Consider the \Sigma-product \displaystyle Z=\Sigma_{\alpha<\omega_1}Z_\alpha about the fixed point p where each Z_\alpha=\omega_1+1=[0,\omega_1]. Based on the observation above, \displaystyle Z=\Sigma_{\alpha<\omega_1}Z_\alpha is homeomorphic to \displaystyle Z_1=\Sigma_{0<\alpha<\omega_1}Z_\alpha \times (\omega_1+1). In turn Z_1 is homeomorphic to \displaystyle Z_2=\Sigma_{\alpha<\omega_1}Z_\alpha \times (\omega_1+1)=Z \times (\omega_1+1). Based on the lemma, Z contains a closed copy of \omega_1. It can be shown that \omega_1 \times (\omega_1+1) is not normal (see a proof in this blog). Thus Z \cong Z \times (\omega_1+1) contains the non-normal closed subspace \omega_1 \times (\omega_1+1) and is thus not normal.

Theorem 1. Let \lbrace{X_\alpha}\rbrace_{\alpha \in S} be a family of compact spaces. Then \displaystyle X=\Sigma_{\alpha \in S}X_\alpha is countably compact.

Proof. Let A \subset X=\Sigma_{\alpha \in S}X_\alpha be a countably infinite set. We show that A has an accumulation point in X. For each f \in A, let T_f \subset S be the countable set on which f \neq the fixed point p. Since A is countable, T=\bigcup_{f \in A}T_f is countable. We can consider A as a subspace of Z=\Pi_{\alpha \in T}X_\alpha. Since Z is compact, A has an accumulation point \text{q} in Z. Extend \text{q} by letting q=p on S-T. It follows that q \in X is an accumulation point of A.

Reference

  1. Gul’ko, S. P., On the properties of sets lying in \Sigma-products, Dokl. Acad. Nauk. SSSR, 237, (1977) 505-508 (in Russian).
  2. Rudin, M. E., \Sigma-products of metric spaces are normal, Preprint, 1977.
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