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could anyone help to show that $[0,1]^{\mathbb{N}}$ with respect to the box topology is not compact? Thank you!

Martin Sleziak
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Jean Carr
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3 Answers3

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There’s absolutely nothing wrong with showing non-compactness of $X = \square_{k=0}^\infty [0,1]$ directly by looking at open covers, but there are other ways as well. For instance:

If $X = {\Large \square}_{k=0}^\infty [0,1]$ were compact, its closed subspace $ {\Large \square}_{k=0}^\infty \{0,1\}$ would be compact, but it’s not hard to show that $ {\Large \square}_{k=0}^\infty \{0,1\}$ is an infinite, closed, discrete set in $X$ and therefore cannot be compact.

Even simpler:

For $n\in\mathbb{N}$ let $x_n \in X$ be the point such that $x_n(n) = 1$ and $x_n(k) = 0$ if $k\ne n$. Now consider the set $A = \{x_n:n\in\mathbb{N}\}$. It’s infinite, so if $X$ were compact, $A$ would have a limit point in $X$. But it’s not hard to show that $A$ is a closed, discrete subset of $X$ and therefore has no limit point in $X$.

Pedro
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Brian M. Scott
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    Just so nobody else gets confused like I did, the square box means "Cartesian product with the box topology", not some kind of missing or mis-rendered character in your browser. – Ted Aug 28 '11 at 06:42
  • @Ted: Sorry about that; I used to work a bit with box products and got used to that notation. – Brian M. Scott Aug 28 '11 at 08:36
  • @BrianM.Scott How to prove that $ {\Large \square}{k=0}^\infty {0,1}$ is closed in $ {\Large \square}{k=0}^\infty [0,1]$? – Twnk May 16 '15 at 03:59
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    @user: For $n,k\in\Bbb N$ let $U_n(k)=(0,1)$ if $n=k$ and $[0,1]$ otherwise. For $n\in\Bbb N$ let $U_n=\prod_{k\in\Bbb N}U_n(k)$. Then $U_n$ is open in ${\Large\square}{k\in\Bbb N}[0,1]$, and $$\left({\Large\square}{k\in\Bbb N}[0,1]\right)\setminus{\Large\square}{k\in\Bbb N}{0,1}=\bigcup{n\in\Bbb N}U_n;.$$ – Brian M. Scott May 16 '15 at 04:26
  • @BrianM.Scott Thanks! Do you think this proof works also for $[0,1]^{[0,1]}$? – Twnk Jun 02 '15 at 06:41
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    @user: $[0,1]^{[0,1]}$ with the box topology? Yes. Or you could you observe that $[0,1]^{[0,1]}$ has a closed subspace homeomorphic to ${\Large\square}_{k\in\Bbb N}[0,1]$, and since the latter is not compact, the former cannot be either. – Brian M. Scott Jun 02 '15 at 15:52
  • @BrianM.Scott Yes, with the box topology. Thank you! – Twnk Jun 02 '15 at 16:23
  • @user: You’re welcome! – Brian M. Scott Jun 02 '15 at 16:31
  • Hello! How do you prove that A is closed? – angushushu Dec 13 '20 at 19:18
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    @Bmo: If $x\notin A$, then either some $x(n)\in(0,1)$, or there are $n\ne m$ such that $x(n)=x(m)=1$, or $x(n)=0$ for every $n\in\Bbb N$. In the first two cases $x$ even has a basic open set in the ordinary product topology that is disjoint from $A$, and in the last case $[0,1)^\Bbb N$ is a basic open nbhd of $x$ in the box topology that is disjoint from $A$. – Brian M. Scott Dec 13 '20 at 19:30
  • @BrianM.Scott thank you so much! That's very clear! Just wondering do you mean there's one of the three conditions for some n∈N? – angushushu Dec 13 '20 at 20:49
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    @Bmo: You’re very welcome. For each $x\notin A$ at least one of the three conditions applies, and if it’s one of the first two conditions, there is at least one $n$ or one pair of $n$ and $m$ for which it applies. – Brian M. Scott Dec 13 '20 at 20:52
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Let $a=(a_1,a_2,\dots)$ be a sequence consisting of only $0$'s and $1$'s and define $U_{a_n}=[0,\frac{3}{4})$ if $a_n=0$ and $U_{a_n}=(\frac{1}{4},1]$ if $a_n=1$. Also, define $U_a=\prod_{n=1}^{\infty} U_{a_n}$.

Exercise 1: Prove that $U_a$ is an open subset of $[0,1]^{\mathbb{N}}$ in the box topology but that it is not an open subset of $[0,1]^{\mathbb{N}}$ in the product topology.

Exercise 2: Prove that the collection of subsets of $[0,1]^{\mathbb{N}}$ of the form $U_a$ where $a=(a_1,a_2,\dots)$ is a sequence consisting of only $0$'s and $1$'s is an open cover of $[0,1]^{\mathbb{N}}$ in the box topology.

Exercise 3: Does this open cover have a finite subcover?

The following problems should be of additional interest:

Problem 1: Let $\{X_n\}_{n\in\mathbb{N}}$ be a collection of topological spaces. If the box topology on $\prod_{n=1}^{\infty} X_n$ is a compact topological space, what can you deduce about the $X_n$'s? In other words, how do the topologies on the individual $X_n$'s affect the compactness of $\prod_{n=1}^{\infty} X_n$ in the box topology?

Problem 2: In the context of Problem 1, if $\prod_{n=1}^{\infty} X_n$ is compact in the box topology, then prove that $X_n$ is compact for all $n\in\mathbb{N}$.

Problem 3: A topological space $X$ is said to be countably compact if every countable open cover of $X$ has a finite subcover. Is $[0,1]^{\mathbb{N}}$ countably compact in the box topology?

I hope this helps!

José Carlos Santos
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Amitesh Datta
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  • Problem 1 is best set in the context of $T_1$-spaces; otherwise a complete answer gets a little messy. Assuming that one already has the Tikhonov theorem, Problem 2 has an easy solution that doesn’t require knowing the result of Problem 1. (Not a complaint, just an observation.) – Brian M. Scott Aug 28 '11 at 08:57
  • @Brian Of course, you are right. I did not suggest that *Problem 2* requires *Problem 1; I only suggested that the notation of Problem 2* is borrowed from *Problem 1* (thus, "In the context of *Problem 1"). Also, I do not think one needs Tikhonov's theorem to solve Problem 2; one can note that the projection map $\prod_{n=1}^{\infty} X_n\to X_i$ is continuous for all $i\in\mathbb{N}$. Problem 1* is more of an open-ended problem; the OP is free to interpret the problem whichever way he/she chooses. If he/she chooses to assume the $T_1$-axiom, then this is fine. – Amitesh Datta Aug 28 '11 at 23:09
  • Hi. I took the liberty to fix the definition of countably compact in Problem 3. – PatrickR Jan 01 '23 at 08:15
  • Thank you so much @PatrickR ! – Amitesh Datta Jan 01 '23 at 08:19
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$$\left\{\displaystyle\prod_{n\in \mathbb{N}} I_n : I\in \left\{[0,\frac23),(\frac13,1]\right\}^\mathbb{N}\right\}$$

is an open cover of $[0,1]^{\mathbb{N}}$ with the box topology that does not have a finite subcover.