Proving $[0,1]^{\omega}$ in box topology is not compact Here is my attempt at proving this, any ideas if it is completely correct?
Let $U^{(0)}=[0,\frac{3}{5}), U^{(1)}=(\frac{2}{5},1]$. Let $x=(x_n)_{n \in \mathbb{N}}$ be an arbitrary sequence of $0's$ and $1's$ in the collection $C$ of all infinite sequences of $0's$ and $1's$. With each sequence $x$ in this collection, associate a sequence of subsets of $[0,1]$ defined by $$U_{nx}= \begin{cases}  U^{(1)}   &\text{ if } x_n=1 \\
          U^{(0)}  & \ \text{if} \ x_n=0\end{cases}$$ Then the collection $S$ of all products of the form $\prod_{n \in \mathbb{N}} U_{nx}$,which are open in $[0,1]^{\omega}$ where $x$ ranges over the set of all binary sequences is an open cover for $[0,1]^{\omega}$ with the box topology.Since for any $k \in \mathbb{N}$ there are at least two elements in $S$ such that the $k$'th factor in the products differ. Now if any element is removed from $S$, one of the infinite sequences $x$ will be removed from $C$, and this sequence will have exactly one coordinate differing from all the other sequences in either a $0$ or $1$ in say the $m$'th slot. So the $m$'th factor in the product of the union of all the elements of this new collection $S'$ will either  be $[0,\frac{3}{5})$ or $(\frac{2}{5},1]$ and hence the collection upon removing the element from $S$ will not cover $[0,1]^{\omega}$. Thus since $S$ is infinite any finite subset of $S$ will certainly not cover the space.
 A: It’s correct but could be stated much more clearly. The key point, which you saw but didn’t really state explicitly, is that for each $x\in C$, the set $\prod_nU_{nx}$ is the only member of $S$ that contains $x$. Your argument for this could be much clearer. Clearly $x\in\prod_nU_{nx}$. If $y\in C\setminus\{x\}$, however, there is an index $k$ such that $y_k\ne x_k$. But then $x_k\notin U_{ky}$, so $x\notin\prod_nU_{ny}$.
$S$ is in fact an irreducible open cover of $[0,1]^\omega$: each member of $S$ contains a point that is not in any other member of $S$, so no proper subset of $S$ covers $[0,1]^\omega$.
Note that essentially the same argument can be used to show that the box product $\{0,1\}^\omega$ is not compact (where $\{0,1\}$ has the discrete topology): here you’re working directly with $C$ as a box product. Since this space is a closed subset of $[0,1]^\omega$, that is another way to see that $[0,1]^\omega$ cannot be compact: if it were, all of its closed subsets would be compact as well.
