TITLE: $[0,1]^{\mathbb{N}}$ with respect to the box topology is not compact
QUESTION [12 upvotes]: could anyone help to show that $[0,1]^{\mathbb{N}}$ with respect to the box topology is not compact? Thank you!

REPLY [14 votes]: 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 $X$.