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I am trying to understand why the following space

$$X=\prod_{n\in \mathbb{N}}[0,1]=[0,1]^{\mathbb{N}}$$

is not compact with respect to the box topology. Specifically, I would like to understand a particular open cover of $X$ (and why it doesn't have a finite subcover) that I have seen given in the answers to another question on this site, but that I have not been able to understand on my own.

In this question, a user gives the following open cover of $X$:

$$\mathscr{F}=\left\{\prod_{n \in \mathbb{N}} I_{n}:I\in \left\{[0,\frac{2}{3}),(\frac{1}{3},1] \right\}^{\mathbb{N}} \right\}.$$

If I understand correctly, it seems that each $I$ is a sequence of numbers whose terms are contained in the set $[0,\frac{2}{3})$ or $(\frac{1}{3},1]$, which means that each member of the collection is a countably infinite product of these sequences? I am having a hard time understanding what a countably infinite product of sequences is exactly and how it covers the space of all sequences on the set $[0,1]$, and why it doesn't have a finite subcover.

As far as I understand, in the box topology, the open sets of $X$ would be of the form

$$\prod_{n\in \mathbb{N}}V_n$$

where each $V_n$ is an open subset of $[0,1]$. And I am guessing that the sets in $\mathscr{F}$ are of this form, but I can't see why. This would seem to me to require that a sequence of points in $[0,1]$ be an open subset of $[0,1]$; though I am probably thinking about this wrong. Any help would be appreciated.

Martin Sleziak
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dwar
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1 Answers1

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Suppose we label $[0 \frac{2}{3} )$ as $V^0$ and $( \frac{1}{3} , 1 ]$ as $V^1$. Now given any infinite binary sequence $\tau = \langle \tau(n) \rangle_{n \in \mathbb{N}}$ we may associate to $\tau$ a sequence $\langle V_n(\tau) \rangle_{n \in \mathbb{N}}$ of these sets as follows: $$V_n ( \tau ) = \begin{cases} V^0, &\text{if }\tau(n) = 0 \\ V^1, &\text{if }\tau(n) = 1. \end{cases}$$ (Somewhat more succinctly, $V_n (\tau) = V^{\tau(n)}$.) Given a fixed $\tau$, it follows that $\langle V_n (\tau) \rangle_{n \in \mathbb{N}}$ is a sequence of open subsets of $[0,1]$, and so their product $$W(\tau) := \prod_{n\in\mathbb{N}} V_n(\tau)$$ is an open subset of ${\large \Box}_{n \in \mathbb{N}} [0,1]$.

It is the family of all these $W(\tau)$ (as $\tau$ ranges over all infinite binary sequences) that comprise the open cover discussed in that answer.

As for what (basic) open sets look like in the box topology: They're (open) boxes! Unrestricted products of open sets in the respective coordinates. In some sense, this makes them somewhat easier to visualise than (basic) open sets in the usual product topology, where we have to add the caveat that only finitely many "sides" of the (open) box not the entire space at that coordinate. (Or, at least less checking is required to show that a supposed basic open set actually is as advertised.)

user642796
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  • That makes sense. So if we remove one of the $W(\tau)$'s from our open cover, since each binary sequence $\tau$ is distinct, there will be some term $n$ in $\tau$ that is different from the corresponding term in all other binary sequences. Then $\langle V_n(\tau)\rangle$ differs from all other such sequences in term $V_n(\tau)$. If, say, this $V_n(\tau)=V^0$, then our cover minus the $W(\tau)$ we removed will not cover sequences on $[0,1]$ whose $i$-th term is an element in $V_i(\tau)$, but whose $n$-th term is an element in $V^0$. So we need all sets in our cover. Does that sound right? – dwar Feb 20 '14 at 18:43
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    @dwar: Yes. To be a bit more explicit, note that each infinite binary sequence $\tau = \langle \tau(n) \rangle_{n \in \mathbb{N}}$ is itself an element of $\prod_{n \in \mathbb{N}} [0,1]$, and we can show that if $\tau \in W ( \sigma )$, then $\sigma = \tau$. – user642796 Feb 20 '14 at 18:50
  • Oh. Yes. That does make things much more concrete. Thanks a lot! – dwar Feb 20 '14 at 21:30