Showing the infinite product space of $[0, 1]^\omega$ is not compact with respect to the box topology using the notion of open covers

Question

I wish to show that the space $[0,1]^\omega$ in the box topology is not compact using the notion of open covers. The box topology has been a topic where I have struggled in my studies of topology, and I am having trouble trying to come up with an open cover that has no finite sub-cover for this space (I understand that there are other ways to prove that this space is not compact, but my goal is to use the notion of open covers and finite sub-covers in the proof). Any help would be greatly appreciated

Answer 1

HINT: For $n\in\omega$ let $x_n$ be the point

$$x_n(k)=\begin{cases} 1,&\text{if }k=n\\ 0,&\text{otherwise} \end{cases}$$

in $X=\square[0,1]^\omega$. For $n\in\omega$ let $F_n=\{x_k:k\ge n\}$, and let $U_n=X\setminus F_n$.

• Show that each $F_n$ is closed. You may find it helpful to begin by shoing that $F_0$ is a closed, discrete subset of $X$.
• Let $\mathscr{U}=\{U_n:n\in\omega\}$; show that $\mathscr{U}$ is an open cover of $X$ with no finite subcover.

Answer 2

Show that the product topology on $[0,1]^\omega$ is strictly coarser than the box topology. Since a compact topology is minimal among T2 topologies, the box topology is therefore not compact.