What is a finite subcover of $[0,1]^{[0,1]}$?

Question

According to Tychonoff's theorem, under the standard topology, $[0,1]^{[0,1]}$ is compact. However, I cannot think of a finite subcover of this space. Also, how does this reconcile with the fact that if $K$ is a compact subset of $(0,1)^{[0,1]}$, then $K$ must only be a finite Cartesian product of compact subset of $(0,1)$? Why cannot $K$ be something like $[0.1,0.5]^{[0,1]}$? Or maybe I am missing something? Thank you very much!

Answer 1

You're right about the final point: in the (non-compact!) space $(0,1)^{[0,1]}$ (in the product topology) the subset $[\frac{1}{10},\frac{1}{2}]^{[0,1]}$ is indeed compact (by the same Tychonoff theorem); it is not true that we can only have a finite product of "compact component spaces"; you seem to be confused with open sets, where basic open sets depend on finitely many coordinates.

There are plenty of finite covers, but the point of compactness is that every open cover of $[0,1]^{[0,1]}$ has a finite subcover.

E.g. $U = \left\{f \in [0,1]^{[0,1]}: f(0)< \frac{3}{4} \right\}$ is open, and so is $V = \left\{f \in [0,1]^{[0,1]}: f(0) > \frac{1}{2} \right\}$ and $U \cup V = [0,1]^{[0,1]}$, and we can do similar things with other finite covers on finitely many coordinates.