$[0,1]^{\mathbb{N}}$ with respect to the box topology is not compact could anyone help to show that $[0,1]^{\mathbb{N}}$ with respect to the box topology is not compact? Thank you!
 A: Let $a=(a_1,a_2,\dots)$ be a sequence consisting of only $0$'s and $1$'s and define $U_{an}=[0,\frac{3}{4})$ if $a_n=0$ and $U_{an}=(\frac{1}{4},1]$ if $a_n=1$. Also, define $U_a=\prod_{n=1}^{\infty} U_{an}$.
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!
A: $$\left\{\displaystyle\prod_{n\in \mathbb{N}} I_n : I\in \left\{[0,\frac23),(\frac13,1]\right\}^\mathbb{N}\right\}$$
is an open cover of $[0,1]^{\mathbb{N}}$ with the box topology that does not have a finite subcover.
A: 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 $X$.
