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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?

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Are you the same user as the one which asked… – Asaf Karagila Dec 25 '11 at 22:53

2 Answers 2

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. $\dashv$

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If $X_i$ are nontrivial, then the answer is no.

The product of compact spaces is compact, as Tychonoff's theorem tells us. It is even separable, since compact metric spaces are separable, and the product of continuum many separable spaces is separable.

From this we have that the product is not first countable, since Hausdorff + separable + first countable implies the cardinality of the space is at most continuum, where as the product of $2^{\aleph_0}$ many sets of at least two points each is much more than the continuum.

This implies that the resulting space is not metrizable, since a metric space is always first countable.

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Can it be sequentially compact? – webster Dec 25 '11 at 21:48
@webster: In the Wikipedia page the examples and counterexamples say that the uncountable product of the unit interval is compact but not sequentially compact. I'd assume that if the spaces are small enough then it might still be sequentially compact, but I'm not sure about that. – Asaf Karagila Dec 25 '11 at 21:51
Then is Martin's answer below correct?… – webster Dec 25 '11 at 21:58
Not even if they’re two-point discrete spaces; I’m writing it up now. – Brian M. Scott Dec 25 '11 at 22:00
@Brian: Interesting! Thanks! – Asaf Karagila Dec 25 '11 at 22:10

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