# 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.