Showing that a countable product of unit intervals is not compact in the box and uniform topology.

Question

Let $I = [0,1]$ and $I^{\omega}$ be the countable product of unit closed interval I, where each $I$ is given the subspace topology of $R$ in the usual topology.

I am trying to show that $I^{\omega}$ is not compact in the box and the uniform topology by showing that there is an open cover without a finite subcover.

In the box topology case, consider the open cover,

${∏_{n∈N} I_n :I_n∈{[0,2/3 ),(1/3 ,1]} }$

It is intuitively clear that there is no finite subcover of this open cover, yet I cannot state it out in words.

For the uniform topology case, I cannot think of such an open cover. Can somebody help me?

Answer 1

Let’s begin by completing your idea for the box topology. Let $U=\left[0,\frac23\right)^\omega$; this is open in the box topology. For $n\in\omega$ let

$$V_n=\left\{\langle x_k:k\in\omega\rangle\in I^\omega:x\in\left(\frac13,1\right]\right\}\;;$$

each $V_n$ is open in the box topology — indeed, even in the ordinary product topology — and $\mathscr{V}=\{U\}\cup\{V_n:n\in\omega\}$ is an open cover of $I^\omega$. Suppose that $\mathscr{V}_0\subseteq\mathscr{V}$ is finite. Let $F=\{n\in\omega:V_n\in\mathscr{V}_0\}$. Define a point $x=\langle x_n:n\in\omega\rangle\in I^\omega$ as follows:

$$x_n=\begin{cases} 0,&\text{if }n\in F\\ 1,&\text{otherwise}\;. \end{cases}$$

Then $$x\notin U\cup\bigcup_{n\in F}V_n\supseteq\bigcup\mathscr{V}_0\;,$$

so $\mathscr{V}_0$ doesn’t cover $I^\omega$, and $\mathscr{V}$ has no finite subcover.

For the uniform topology I think that another approach is easier. For $n\in\omega$ define the point $x^{(n)}=\langle x_k^{(n)}:k\in\omega\rangle\in I^\omega$ by

$$x_k^{(n)}=\begin{cases} 1,&\text{if }k=n\\ 0,&\text{otherwise}\;. \end{cases}$$

It’s not hard to show that $A=\{x^{(n)}:n\in\omega\}$ is an infinite subset of $I^\omega$ with no limit point in the uniform topology, which is enough to show that $I^\omega$ is not compact in the uniform topology. If you want to use open covers, for each $n\in\omega$ let $B_n$ be the open $\frac12$-ball centred at $x^{(n)}$. (In other words, $B_n=\prod_{n\in\omega}I_k$, where $I_k=\left[0,\frac12\right)$ if $k\ne n$, and $I_n=\left(\frac12,1\right]$.) Then show that $A$ is closed, and let $U=I^\omega\setminus A$; $\mathscr{U}=\{U\}\cup\{B_n:n\in\omega\}$ is then an open cover of $I^\omega$ with no finite subcover, since for each $n\in\omega$ the only member of $\mathscr{U}$ containing $x^{(n)}$ is $B_n$.

Answer 2

One way to do it: consider the subset $A = \{0,1\}^\omega$ consisting of sequences whose entries are all either 0 or 1. Show that in each topology, $A$ is closed. For each $x \in A$, find an open set $U_x$ with $U_x \cap A = \{x\}$. Show that the open cover $\{A^c\} \cup \{U_x : x \in A\}$ has no finite subcover.