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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 http://math.stackexchange.com/questions/85783/constructing-a-convergent-subsequence2011-12-25

2 Answers 2

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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

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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    What was $φ$ and what did $B(φ)$ and $B(F)$ mean in the last part, since $B$ has only be defined with indices over $\omega_1$ before? Thanks.2017-09-26
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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?2011-12-25
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    @webster: In [the Wikipedia page](http://en.wikipedia.org/wiki/Sequentially_compact_space) 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.2011-12-25
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    Then is Martin's answer below correct? http://math.stackexchange.com/questions/85783/constructing-a-convergent-subsequence2011-12-25
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    Not even if they’re two-point discrete spaces; I’m writing it up now.2011-12-25
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    @Brian: Interesting! Thanks!2011-12-25
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    The original question is: suppose for each $i \in [0,1]$ we have a sequence $\{X_{i,n}\}$. Suppose the set of possible $X_{i,n}$ is sequentially compact. Can I construct a subsequence $n'$ such that for each $i$, $X_{i,n'} \rightarrow X_i$ for some $X_i$?2011-12-25
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    In Willard's general topology, it is claimed that a product of $\le\aleph_1$ sequentially compact spaces is seqentially compact (Scarborough and Stone, "Products of Nearly Compact Spaces", Trans. Amer. Math Soc, 124., 131-147 (1966) ). But, assuming the continuum hypothesis, an uncountable product of $T_1$-spaces, each having more than one point, is never sequentially compact. (see exercise 17G and the notes)2011-12-25
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    I see. Then is there any condition on the set of possible $X_{i,n}$ that will allow me to construct a subsequence $n'$ in the above setting?2011-12-25
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    @webster I'm not sure... I'm awaiting Brian's answer.2011-12-25
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    I take that back: I misremembered. For any regular cardinals $\omega_1\le\kappa\le\lambda$ it’s consistent that $2^\omega=\lambda$ and $\kappa$ is the smallest cardinal such that $\{0,1\}^\kappa$ is not sequentially compact. See Theorem 5.1 in Eric K. van Douwen, *The Integers and Topology*, in *The Handbook of Set-Theoretic Topology*, K. Kunen & J.E. Vaughan, eds.2011-12-25
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    Thank you. So am I right in saying that it is impossible in case of uncountable products, although it is possible with countable products?2011-12-25
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    @David: Willard’s comment in the notes is simply wrong; what S&S prove is that the product of at most $\omega_1$ sequentially compact spaces is *countably* compact, as Willard mentions in 17G.2011-12-25
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    @Brian M. Scott Not the only false statement in the book. (It's my favorite Topology text, though. I need to get the second edition; hopefully the errors there have been corrected...)2011-12-25