TITLE: Cartesian product of uncountable number of compact metric spaces
QUESTION [11 upvotes]: Suppose $ i \in [0,1]$ and for each $i$, $X_i$ is a compact metric space. Then, is it that a Cartesian product of $X_i$ over $ i \in [0,1]$ is also a compact metric space?

REPLY [14 votes]: Since the subject came up, here’s the basic story on sequential compactness in such products.
Let $[\omega]^\omega$ denote, as usual, the set of infinite subsets of $\omega$. A family $\mathscr{S}\subseteq[\omega]^\omega$ is called a splitting family if it has the following property:

$\qquad\quad$for each infinite $A\subseteq\omega$ there is an $S\in\mathscr{S}$ such that $|A\cap S|=|A\setminus S|=\omega$.

The minimum cardinality of such a family is denoted by $\mathfrak{s}$. This splitting cardinal $\mathfrak{s}$ is intimately connected with sequential compactness of products.

Prop. 1: Suppose that $X=\displaystyle\prod_{\xi<\mathfrak{s}}X_\xi$, where each $X_\xi$ is a Hausdorff space with at least two points; then $X$ is not sequentially compact.
Proof: Let $\mathscr{S}=\{S_\xi:\xi<\mathfrak{s}\}$ be a splitting family. For $\xi<\mathfrak{s}$ fix distinct points $x_0^\xi,x_1^\xi\in X_\xi$. For $n\in\omega$ define $y_n\in X$ by $$y_n(\xi)=\begin{cases}
x_1^\xi,&\text{if }n\in S_\xi\\\\
x_0^\xi,&\text{if }n\notin S_\xi\;.
\end{cases}$$ For any $A\in[\omega]^\omega$ there is a $\xi<\mathfrak{s}$ such that $|A\cap S_\xi|=|A\setminus S_\xi|=\omega$; then clearly $\langle y_k(\xi):k\in A\rangle$ doesn’t converge in $X_\xi$, so $\langle y_k:k\in\omega\rangle$ has no convergent subsequence in $X$. $\dashv$

In the other direction we have:

Prop. 2: Suppose that $X=\displaystyle\prod_{\xi<\kappa}X_\xi$, where $\kappa<\mathfrak{s}$ and each $X_\xi$ is a compact Hausdorff space of weight at most $\kappa$; then $X$ is sequentially compact.
Proof: Let $\langle y_n:n\in\omega\rangle$ be a sequence in $X$. $X$ has a base $\mathscr{B}$ of cardinality $\kappa$; Let $$\mathscr{B}\,'=\big\{B\in\mathscr{B}:|\{n\in\omega:y_n\in B\}|=\omega\big\}\;,$$ the set of $B\in\mathscr{B}$ containing infinitely many terms of the sequence. $\mathscr{B}\,'$ is not a splitting family, since $|\mathscr{B}\,'|\le\kappa<\mathfrak{s}$, so there is an infinite $A\subseteq\omega$ such that for every $B\in\mathscr{B}\,'$, one of the sets $\{n\in A:y_n\in B\}$ and $\{n\in A:y_n\notin B\}$ is finite. $X$ is compact, so $\{y_n:n\in A\}$ has a cluster point $y$, and I claim that $\langle y_n:n\in A\rangle\to y$. To see this, note that if $B\in\mathscr{B}$ is a nbhd of $y$, then $B\in\mathscr{B}\,'$ and hence $\{n\in A:y_n\notin B\}$ is finite (since $\{n\in A:y_n\in B\}$ is infinite by the choice of $y$). Thus, every nbhd of $y$ contains a tail of $\{y_n:n\in A\}$, which therefore converges to $y$. $\dashv$

In particular, and of most interest here:

Cor. 3: Suppose that $X=\displaystyle\prod_{\xi<\kappa}X_\xi$, where each $X_\xi$ is a compact metric space with at least two points; then $X$ is sequentially compact iff $\kappa<\mathfrak{s}$. $\dashv$

This obviously raises the question of just how big $\mathfrak{s}$ is. Clearly $[\omega]^\omega$ is a splitting family, so $\mathfrak{s}\le\mathfrak{c}=2^\omega$. On the other hand, $\mathfrak{s}$ must be uncountable:

Prop. 4: $\mathfrak{s}>\omega$.
Proof: $\mathscr{C}=\{C_n:n\in\omega\}\subseteq[\omega]^\omega$ be arbitrary. Let $X_0=\mathscr{A}_0=\mathscr{B}_0=\varnothing$. Suppose that $n\in\omega$ and that $\{\mathscr{A}_n,\mathscr{B}_n\}$ is a partition of $\{C_k:k<n\}$ such that $\bigcap\mathscr{A}_n$ is infinite (where we define $\bigcap\varnothing=\omega$). If $C_n\cap\bigcap\mathscr{A}_n$ is infinite, let $$\begin{align*}
&\mathscr{A}_{n+1}=\mathscr{A}_n\cup\{C_n\},\\
&\mathscr{B}_{n+1}=\mathscr{B}_n\,,\text{ and}\\
&X_{n+1}=X_n\cup\left\{\inf \Big(C_n\cap\bigcap\mathscr{A}_n\setminus X_n\Big)\right\}\;.
\end{align*}$$ Otherwise, let $$\begin{align*}
&\mathscr{A}_{n+1}=\mathscr{A}_n\,,\\
&\mathscr{B}_{n+1}=\mathscr{B}_n\cup\{C_n\},\text{ and}\\
&X_{n+1}=X_n\cup\left\{\inf \Big(\bigcap\mathscr{A}_n\setminus (C_n\cup X_n)\Big)\right\}\;.
\end{align*}$$ Let $X=\displaystyle\bigcup_{n\in\omega}X_n$, $\mathscr{A}=\displaystyle\bigcup_{n\in\omega}\mathscr{A}_n$, and $\mathscr{B}=\displaystyle\bigcup_{n\in\omega}\mathscr{B}_n$; $\{\mathscr{A},\mathscr{B}\}$ is a partition of $\mathscr{C}$, $X\setminus A$ is finite for every $A\in\mathscr{A}$, and $X\cap B$ is finite for every $B\in\mathscr{B}$, so $\mathscr{C}$ doesn’t split $X$. $\dashv$

Thus, $\omega_1\le\mathfrak{s}\le\mathfrak{c}$. Under $\text{CH}$, therefore, the product of $\omega_1$ non-trivial compact metric spaces is not sequentially compact. It’s not hard to show, however, that $\text{MA}+\lnot\text{CH}$ implies that $\mathfrak{s}=\mathfrak{c}$ and hence that the product of $\omega_1$ compact metric spaces is sequentially compact. The proof is similar to that of Prop. 4.

Prop. 5: ($\text{MA}+\lnot\text{CH}$) $\mathfrak{s}=\mathfrak{c}$.
Sketch of Proof: Let $\mathscr{S}\subseteq[\omega]^\omega$ with $|\mathscr{S}\,|<\mathfrak{c}$. Let $$\mathscr{P}=\left\{\langle F,\mathscr{A},\mathscr{B}\,\rangle\in[\omega]^{<\omega}\times[\mathscr{S}\,]^{<\omega}\times[\mathscr{S}\,]^{<\omega}:\left|\bigcap\mathscr{A}\setminus\bigcup\mathscr{B}\right|=\omega\right\}\;.$$ For $\langle F_0,\mathscr{A}_0,\mathscr{B}_0\rangle,\langle F_1,\mathscr{A}_1,\mathscr{B}_1\rangle\in\mathscr{P}$ write $$\langle F_0,\mathscr{A}_0,\mathscr{B}_0\rangle \preceq \langle F_1,\mathscr{A}_1,\mathscr{B}_1\rangle$$ iff $F_1\subseteq F_0$, $\mathscr{A}_1\subseteq\mathscr{A}_0$, $\mathscr{B}_1\subseteq\mathscr{B}_0$, $F_0\setminus F_1\subseteq S$ for every $S\in\mathscr{A}_1$, and $S\cap F_0\subseteq F_1$ for every $S\in\mathscr{B}_1$. It’s routine to verify that $\langle \mathscr{P},\preceq\rangle$ is a ccc partial order in which the sets $$\mathscr{D}_S=\{\langle F,\mathscr{A},\mathscr{B}\,\rangle\in\mathscr{P}:S\in\mathscr{A}\cup\mathscr{B}\}\text{ for }S\in\mathscr{S}$$ and $$\mathscr{D}_n=\left\{\langle F,\mathscr{A},\mathscr{B}\,\rangle\in\mathscr{P}:\forall S\in\mathscr{A}\,\big(|S\cap F\,|\ge n\big)\right\}\text{ for }n\in\omega$$ are dense and open. Let $\mathscr{F}\subseteq\mathscr{P}$ be a filter meeting each of these dense sets. Let $$\begin{align*}X&=\bigcup\left\{F\in[\omega]^{<\omega}:\exists\mathscr{A},\mathscr{B}\in[\mathscr{S}\,]^{<\omega}\big(\langle F,\mathscr{A},\mathscr{B}\rangle\in\mathscr{F}\,\big)\right\},\\
\mathscr{S}_0&=\bigcup\left\{\mathscr{A}\in[\mathscr{S}\,]^{<\omega}:\exists F\in[\omega]^{<\omega}\;\exists\mathscr{B}\in[\mathscr{S}\,]^{<\omega}\big(\langle F,\mathscr{A},\mathscr{B}\rangle\in\mathscr{F}\,\big)\right\},\text{ and}\\
\mathscr{S}_1&=\bigcup\left\{\mathscr{B}\in[\mathscr{S}\,]^{<\omega}:\exists F\in[\omega]^{<\omega}\;\exists\mathscr{A}\in[\mathscr{S}\,]^{<\omega}\big(\langle F,\mathscr{A},\mathscr{B}\rangle\in\mathscr{F}\,\big)\right\}.\\
\end{align*}$$ Then $\{\mathscr{S}_0,\mathscr{S}_1\}$ is a partition of $\mathscr{S}$, $X\in[\omega]^\omega$, $|X\subseteq S|<\omega$ for every $S\in\mathscr{S}_0$, and $|X\cap S|<\omega$ for every $S\in\mathscr{S}_1$, so $\mathscr{S}$ doesn’t split $X$. $\dashv$

There are many other consistency results, but this is already enough to show that much depends on your set theory.
And as long as I’ve written this much already, here’s a direct proof that the product of uncountable many non-trivial compact metric spaces is not first countable.

Let $Y$ be such a space. Then $Y$ contains a copy of $X=D^{\omega_1}$, where $D$ is the two-point discrete space, so it suffices to show that $X$ is not first countable. Let $z$ be the element of $X$ such that $z(\xi)=0$ for all $\xi\in\omega_1$. For each $\xi\in\omega_1$ let $B_\xi=\{x\in X:x(\xi)=0\}$. If $\mathscr{U}=\{U_n:n\in\omega\}$ is a countable local base at $z$, then for each $\xi\in\omega_1$ there is an $n(\xi)\in\omega$ such that $U_{n(\xi)}\subseteq B_\xi$. For $k\in\omega$ let $A_k=\{\xi\in\omega_1:n(\xi)=k\}$; there is some $k\in\omega$ such that $|A_k|=\omega_1$. Now choose $B(\varphi)\in\mathscr{B}$ such that $z\in B(\varphi)\subseteq U_k$, let $F=\operatorname{dom}\,\varphi$, and choose $\eta\in A_k\setminus F$. Define $x\in X$ by $$x(\xi)=\begin{cases}1,&\text{if }\xi=\eta\\0,&\text{otherwise}\;;\end{cases}$$ then $x\in B(F)\setminus B_\eta\subseteq B(F)\setminus U_k$, which is a contradiction. $\dashv$